# Brilliant Inequality Contest - Season 1

Welcome all to the first ever Brilliant Inequality Contest. Like the Brilliant Integration Contest, the aim of the Inequality Contest is to improve skills and techniques often used in Olympiad-style Inequality problems. But above all, the main reason is so we can have fun.

Anyone is allowed to participate, as long as they adhere to the following rules.

1. I will post the first problem. If someone solves it, he or she can post a solution and then must post a new problem.

2. A solution must be posted below the thread of the problem. Then, the solver must post a new problem as a separate thread.

3. Please make a substantial comment. Spam or unrelated comments will be deleted. (To help me out, try to delete your comments within an hour of when you posted them, if they are not the solutions.)

4. Make sure you know how to solve your own problem before posting it, in case no one else is able to solve it within 48 hours. Then, you must post the solution and you have the right to post a new problem.

5. If the one who solves the last problem does not post a new problem in 24 hours, the creator of the previous problem has the right to post another problem.

6. The scope of the problems is Olympiad-style inequalities.

7. You are not allowed to post problems requiring calculus in the solutions (use of differentiation to prove a curve is concave or convex is allowed).

8. Lagrange Multipliers are not allowed to be used in a solution.

9. Inequalities allowed to be used are AM-GM, Muirhead, Power Mean and Weighted Power Mean, Cauchy-Schwarz, Holder, Rearrangement, Chebyshev, Schur, Jensen, Karamata, Reverse Rearrangement and Titu's lemma.

10. Try to post the simplest solution possible. For example, if someone posted a solution using Holder, Titu's and Cauchy, when there is a solution using only AM-GM, the latter is preferred.

SOLUTION OF PROBLEM (insert problem number here)

PROBLEM (insert problem number here)

Remember to reshare this note so it goes to everyone out there. And above all else, have fun!!!

PROBLEM 1

Let $x$, $y$ and $z$ be positive reals such that $x+y \geq z$, $y+z \geq x$ and $z+x \geq y$. Find families of solutions for $(x, y, z)$ such that the following inequality is satisfied.

$2x^2 (y+z) + 2y^2 (z+x) + 2z^2 (x+y) \geq x^3 + y^3 + z^3 + 9xyz$

P.S.: For those who want to discuss problem solutions, they can do so here.

Note by Sharky Kesa
5 years, 3 months ago

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@Chris Galanis, @Sualeh Asif, @Daniel Liu, @Sharky Kesa, @Harsh Shrivastava, @Svatejas Shivakumar. Thanks everyone fit making this such a beautiful contest, hope it went on! For season 2 I will be the organiser, comment please for it.

- 5 years, 1 month ago

Waiting forit to begin...

- 5 years, 1 month ago

More than welcome :)

- 5 years, 1 month ago

Sure!

- 5 years, 1 month ago

@Sharky Kesa The contest ended? Wish it could have continued for some for time. Anyway thanks to everyone who participated in this contest, learnt a lot and had a lot of fun.

- 5 years, 1 month ago

Problem 28

For the real numbers $a, b, c, d$ prove that: $-1\le \dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1$

- 5 years, 2 months ago

For the right hand side note that for positive real numbers(or all negative) $(a^2 +b^2)(c^2+d^2)\geq (ac+bd)^2$ By C.S .

Now it suffices to consider the case when one of $a,b,c,d$ are negative. If one of them is negative. WLOG assume $a$ then let $A=-a$. Thus $(bd-Ac)^2\leq (bd+ac)^2 \leq (a^2+b^2)(c^2+d^2)$ equality holds when $A=a=0$

If any two of them are negative we have two cases .

WLOG assume $a,b$. Then let $A=-a$, $B=-b$, $\therefore (Ac+Bd)^2=(ac+bd)^2$ Otherwise assume both of $a,c$ are negative.Then there multiple is positive and we are done.

The case when 3 are negative is equivalent to the case when one is negative.

- 5 years, 2 months ago

Yet that is not a proof since $a, b, c, d$ may not be positive

- 5 years, 2 months ago

Of course completed the solution.

I tend to keep posting my progress because the mobile app gets stuck

- 5 years, 2 months ago

Problem 27

Given positive reals $a,b,c$, prove that $3\sum_{cyc}\dfrac{a+b}{c}\ge 10+8\cdot \dfrac{a^2+b^2+c^2}{ab+bc+ca}$

- 5 years, 2 months ago

(1) $(a+b)(b+c)(c+a) \ge \dfrac{8}{9}(a+b+c)(ab+bc+ca)$ which is true since by expanding and simplifying we get $\displaystyle \sum_{sym} a^2b \ge 6abc$ which is rearrangement inequality.

(2) $\Big(\displaystyle \sum_{sym} a^2b\Big)+2abc = (a+b)(b+c)(c+a)$ and $(a+b+c)^2 = a^2+b^2+c^2 +2(ab+bc+ca)$

$3 \displaystyle \sum_{cyc} \dfrac{a+b}{c} \ge 10 + 8\cdot \dfrac{a^2+b^2+c^2}{ab+bc+ca}$ Adding $6$ to both sides and using (2) we get:

$3\dfrac{(a+b)(b+c)(c+a)}{abc} \ge 8\dfrac{(a+b+c)^2}{ab+bc+ca}$ Thus by (1) it suffices to prove that:

$(ab+bc+ca)^2\ge 3abc(a+b+c)$ Which is true since by expanding and simplifying both sides we get rearrangement inequality. Hence proved!

- 5 years, 2 months ago

Problem 26

Given positive reals $x$, $y$, $z$ greater than or equal to 1 such that

$\dfrac {1}{x} + \dfrac {1}{y} + \dfrac {1}{z} = 2$

Prove

$\sqrt{x+y+z} \geq \sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1}$

- 5 years, 2 months ago

Another solution:

The problem is equivalent to $(x+y+z)\left(3-\dfrac{1}{x}-\dfrac{1}{y}-\dfrac{1}{z}\right)\ge \left(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\right)^2$

Confirm that $1-\dfrac{1}{x}\ge 0\implies x\ge 1$; thus, by C-S we are done.

- 5 years, 2 months ago

Problem 25

For the positive real numbers $a, b, c$ prove that: $\dfrac{1}{b(a+b)} + \dfrac{1}{c(b+c)} + \dfrac{1}{a(c+a)} \geq \dfrac{27}{2(a+b+c)^2}$

- 5 years, 2 months ago

Solution to Problem 25

This solution is quite long and extremely bashy. Sorry. :P

We have to prove

$\displaystyle \sum_{\text{cyc}} \dfrac {1}{b(a+b)} \geq \dfrac {27}{2(a+b+c)^2}$

Multiplying out the denominators and simplifying (here is the bash, but I have suppressed it), we get

$\displaystyle \sum_{\text{cyc}} 2a^5 b + 6a^4 b^2 + 2a^2 b^4 + 6a^3 b^3 + 8a^4bc \geq \displaystyle \sum_{\text{cyc}} 7a^3 b^2 c + 9a^3 b c^2 + 8 a^2 b^2 c^2$

By AM-GM, we have

$\dfrac {a^5 b + a^4 b^2 + c^4 a^2 + a^2 b^4 + a^2 b^4 + a^3 b^3 + a^3 c^3}{7} \geq \sqrt[7]{a^{21}b^{14}c^7}$

$\dfrac {a^5 b + a^4 b^2 + c^4 a^2 + c^4 a^2 + c^4 a^2 + a^3 b^3 + a^3 b^3 + a^3 c^3 + a^3 c^3}{9} \geq \sqrt[9]{a^{27}b^{9}c^{18}}$

$\dfrac {8a^4bc + 8ab^4c + 8abc^4}{3} \geq \sqrt[3]{24a^6 b^6 c^6}$

These inequations imply

$a^5 b + a^4 b^2 + c^4 a^2 + 2a^2 b^4 + a^3 b^3 + a^3 c^3 \geq 7a^3 b^2 c$

$a^5 b + a^4 b^2 + 3c^4 a^2 + 2a^3 b^3 + 2a^3 c^3 \geq 9a^3 b c^2$

$8a^4bc + 8ab^4c + 8abc^4 \geq 24a^2 b^2 c^2$

Adding the first two equations cyclicly, then adding the third equation, we get the statement we wished to prove. Thus proven!

PS: It took me hours to get the right grouping of terms for the AM-GM.

- 5 years, 2 months ago

You could have grouped them by triples of powers that majorize each other, and then applied AM-GM... Saves you all the time of finding your own grouping ;)

- 5 years, 2 months ago

Sorry for the delay guys, and the inconvenience caused by me.

Problem 24:

Let a,b,c be positive reals.

Prove that $\displaystyle \prod_{cyc} (1 + \dfrac{a}{b}) \geq 2(1 + \dfrac{a+b+c}{ (abc)^{\frac{1}{3}}})$

- 5 years, 3 months ago

Open up the brackets and simplify a bit to get $\large \sum_{cyc} (\frac{a}{b} + \frac{a}{c}) \geq 2\dfrac{(a+b+c)}{(abc)^{\frac{1}{3}}}$

Substitute $a = x^{3}, b= y^{3} and c = z^{3}$, we get $\large \sum_{cyc} (\frac{x^{3}}{y^{3}} + \frac{x^{3}}{z^{3}}) \geq 2\dfrac{(x^{3} + y^{3}+ z^{3})}{xyz} = \sum_{cyc} (\dfrac{x^{2}}{yz} + \dfrac{x^{2}}{yz})$

Now using re-arrangement inequality, we can get our desired result.

- 5 years, 3 months ago

Hmm sorry guyz I won't be able to post solution now because I am a bit busy. Will post a solution by tmmrw evening(my time zone).

- 5 years, 3 months ago

I think this is equivalent to $(1+\dfrac{a}{b})(1+\dfrac{b}{c})(1+\dfrac{c}{a}) \geq 2(1+\dfrac{a+b+c}{(abc)^{1/3}})$

By applying Holder's Inequality to LHS it suffices to prove that $\Big((1+(\dfrac{a}{b} \dfrac{b}{c} \dfrac{c}{a})^{1/3}\Big)^3 \geq 2(1+\dfrac{a+b+c}{(abc)^{1/3}}) \\ \Rightarrow 4 \geq 1+\dfrac{a+b+c}{(abc)^{1/3}} \\ \Rightarrow (abc)^{1/3} \geq \dfrac{a+b+c}{3}$ But by AM-GM this is true only for $a=b=c=1$

Edit: If the problem had the direction of the inequality flipped then we would have to prove that

\begin{aligned} (1+\dfrac{a}{b})(1+\dfrac{b}{c})(1+\dfrac{c}{a}) & \leq 2(1+\dfrac{a+b+c}{(abc)^{1/3}}) \\ & \stackrel{AM-GM}{\leq}2(1+\dfrac{a+b+c}{{a+b+c}/_3}) = 8\end{aligned}

Then it suffices to prove that \begin{aligned} & (a+b)(b+c)(c+a) \leq 8abc \\ & \stackrel{AM-GM}{\leq} \Big(\dfrac{2}{3}(a+b+c)\Big)^3 \leq 8abc \\ & \Rightarrow a^3 + b^3 +c^3 +3a^2b+3a^2c+3ab^2+3ac^2+3b^2c+3bc^2 \leq 21abc \end{aligned} But again that would be true by rearrangement inequality if the direction of the inequality was flipped...

- 5 years, 3 months ago

The problem is that every time you apply an inequality, the inequality weakens a bit. If you apply inequalities in the incorrect fashion or too much, then the original inequality will weaken to the point that it is false.

- 5 years, 2 months ago

- 5 years, 2 months ago

I request someone to post a problem on my behalf. Thanks!

- 5 years, 2 months ago

@Chris Galanis I hope you post your problem soon!

- 5 years, 3 months ago

This was not a solution... I wait for Harsh or someone else to post the solution

- 5 years, 3 months ago

I believe that it isn't a solution. The sign must be interchanged. Even I got that after applying A.M-G.M. You pretty much cannot use anything I guess. Such a strange inequality.

- 5 years, 3 months ago

Since the inequality is homogenous, we assume $abc=1$.
We can factor it to $\sum_{cyc} (a+b)(ab-1)\geq 0$

- 5 years, 3 months ago

Problem 23

Find the greatest positive real number $k$, for which is true that $\dfrac{xy}{\sqrt{(x^2+y^2)(3x^2+y^2)}} \leq \dfrac{1}{k}$ for all positive real numbers $x, y$.

- 5 years, 3 months ago

- 5 years, 3 months ago

Problem 22

Here is a very nice problem I came across.

Given that $a,b,c$ are non-negative reals, prove that $(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2$

- 5 years, 3 months ago

Solution of Problem 22

By Cauchy we got: $(a^2+2)(1+\frac{(b+c)^2}{2}) \geq (a+b+c)^2$

Then it suffices to prove that $(b^2+2)(c^2+2) \geq 3(1+\frac{(b+c)^2}{2})$

By rearranging and simplifying we get: $(b^2-2bc+c^2) + (2b^2c^2-4bc+1) \geq 0 \\ (b-c)^2 + 2(bc-1)^2 \geq 0$ Which is true. Equality holds iff $a=b=c=1$.

- 5 years, 3 months ago

Can you please elaborate on your first step? I don't seem to see the Cauchy. Did you perhaps mean $\dfrac{(b+c)^2}{2}$ instead of $\dfrac{b+c}{2}$?

EDIT: I tried to solve the problem using your idea and it appears that you have only just typoed. Splendid solution!

- 5 years, 3 months ago

Sorry, I fixed that !

- 5 years, 3 months ago

Problem 21

If for the real numbers $a, b, c$ where $bc \neq 0$ is true that $\frac{1-c^2}{bc} \geq 0$, prove that: $10(a^2+b^2+c^2-bc^3) \geq 2ab + 5ac$

- 5 years, 3 months ago

If $|c| \ge 1$ and $\text{sgn}(b) = -\text{sgn}(c)$, then the inequality is $10(a^2+b^2+c^2-bc^3)\ge 10(a^2+b^2+c^2)\ge 2ab+5ac$$\iff (a-b)^2+\dfrac{5}{2}(a-c)^2+\dfrac{13}{2}a^2+9b^2+\dfrac{15}{2}c^2\ge 0$ solved.

If $|c| \le 1$ and $\text{sgn}(b) = \text{sgn}(c)$, then the inequality is $10(a^2+b^2+c^2-bc^3)\ge 10(a^2+b^2+c^2-bc)\ge 2ab+5ac$$\iff (a-b)^2+\dfrac{5}{2}(a-c)^2+5(b-c)^2+\dfrac{13}{2}a^2+4b^2+\dfrac{5}{2}c^2\ge 0$ done.

Not a very strong inequality; no equality case in fact.

- 5 years, 3 months ago

Can this problem be done without calculus? I have no idea how to proceed with this problem.

- 5 years, 3 months ago

Me and Sharky were discussing about it and it showed this can be nearly solved using $x^2 \ge 0, x\in \mathbb{R}$.

- 5 years, 3 months ago

Problem 20

Let $a$, $b$ and $c$ be positive reals such that $abc=8$. Prove that

$\dfrac {a^2}{\sqrt{(1+a^3)(1+b^3)}} + \dfrac {b^2}{\sqrt{(1+b^3)(1+c^3)}} + \dfrac {c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \dfrac {4}{3} .$

- 5 years, 3 months ago

Solution of Problem 20

Note that $\sqrt{1+k^3} = \sqrt{(1+k)(1-k+k^2)} \stackrel{AM-GM}{\leq} \frac{k^2-k+1+k+1}{2} = \frac{k^2+2}{2} \Leftrightarrow \frac{2}{k^2+2} \leq \frac{1}{\sqrt{(1+k^3)}} (*)$ Hence: \begin{aligned} \displaystyle \sum_{cyc} \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} & \geq \frac{4}{3} \\ \stackrel{(*)}{\Rightarrow} \sum_{cyc} \frac{4a^2}{(a^2+2)(b^2+2)} & \geq \frac{4}{3} \\ \Rightarrow 3 \sum_{cyc} a^2(c^2+2) & \geq (a^2+2)(b^2+2)(c^2+2) \\ \Rightarrow \sum_{cyc} 2a^2+ \sum_{cyc} a^2b^2 & \geq 72 \end{aligned} Which is true by applying AM-GM in both sums separately. Hence proved.

- 5 years, 3 months ago

Problem 19 :

Let $x,y,z$ be positive real numbers.Prove that $\displaystyle \sum_{cyc} \dfrac{1}{x^{3} + y^{3} + xyz} \leq \dfrac{1}{xyz}$

- 5 years, 3 months ago

Solution to Problem 19

Since this inequality is homogenous, WLOG $xyz = 1$. We will prove that

$\dfrac {1}{x^3+y^3+1} \leq \dfrac {z}{x+y+z}$

$(x^3+y^3)z+z \geq x+y+z$

Note that $x^3+y^3 = (x+y)(x^2-xy+y^2)$.

$(x+y)(x^2-xy+y^2)z \geq x+y$

$(x^2 + y^2 - xy)z \geq 1$

By AM-GM, $x^2 + y^2 \geq 2xy$. Substituting, we get

$(2xy-xy)z \geq 1$

$xyz \geq 1$

Which is obviously true by our assumption. Thus, proven. The inequality in the question can be achieved by adding cyclicly.

We have already proven (with WLOG $xyz=1$ because of the equation's homogeneity)

$\dfrac {1}{x^3 + y^3 + 1} \leq \dfrac {z}{x+y+z}$

This means that

$\dfrac {1}{x^3 + y^3 + 1} + \dfrac {1}{y^3 + z^3 + 1} + \dfrac {1}{z^3 + x^3 + 1} \leq \dfrac {x+y+z}{x+y+z}$

$\displaystyle \sum_{\text{cyc}} \dfrac {1}{x^3 + y^3 + 1} \leq 1$

Note that since $xyz=1$, $\dfrac {1}{xyz} = 1$. Thus,

$\displaystyle \sum_{\text{cyc}} \dfrac {1}{x^3 + y^3 + xyz} \leq \dfrac {1}{xyz}$

- 5 years, 3 months ago

PROBLEM 18

For positive real numbers $a,b,c$ with $a+b+c=abc$, prove that $\dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}}+\dfrac{1}{\sqrt{1+c^2}} \leq \dfrac{3}{2}$.

- 5 years, 3 months ago

Solution Of Problem 18 :

Substitute a= tanx , b= tany , c= tanz such that $x+y+z= \pi$

Since a,b,c are positive reals, x,y,z are all less than $\frac{\pi}{2}$

Now we want to prove that $\cos x + \cos y + \cos z \leq 1.5$

By Jensen's Inequality,

$\cos \left(\dfrac{x+y+z}{3}\right)\ge \dfrac{\cos x+\cos y+\cos z}{3}$

$\implies \cos x + \cos y + \cos z \leq \dfrac{3}{2}$

Note that $f(p) = \cos p$ is concave down down for all $0 \leq p \leq \frac{\pi}{2}$.This can be checked by differentiating f(p) twice.

- 5 years, 3 months ago

Problem 17

For positive reals $x$, $y$ and $z$ such that $xyz=1$, prove that

$\dfrac {1}{x^3(y+z)} + \dfrac {1}{y^3(z+x)} + \dfrac {1}{z^3 (x+y)} \geq \dfrac {3}{2}$

- 5 years, 3 months ago

This is a very famous inequality that appeared in IMO in the late 90s I believe. My approach goes as follows.

SOLUTION OF PROBLEM 17

Dividing the numerator and denoninator by $x^2,y^2$ and $z^2$ respectively and applying Titu's Lemma we get $\dfrac {1}{x^3(y+z)} + \dfrac{1}{y^3(z+x)}+\dfrac{1}{z^3 (x+y)} \geq \dfrac{(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2}{2(xy + yz+ zx)}=\dfrac{(xy+yz+zx)^2}{2(xy+yz+zx)}=\dfrac{xy+yz+zx}{2}$

Now, by A.M-G.M inequality $\dfrac{xy+yz+zx}{2} \geq 3 \cdot \dfrac{\sqrt[3]{(xyz)^2}}{2}=\dfrac{3}{2}$.

Equality holds if and only if $x=y=z=1$. Hence proved.

- 5 years, 3 months ago

Problem 16

For positive reals $a,b,c$, prove $\frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1.$

Source: ISL

- 5 years, 3 months ago

Solution to Problem 16

I used Jensen's Inequality to solve this problem.

Apply Jensen's on the convex function

$f(x) = \dfrac {1}{\sqrt{x}}$

and weights $a$, $b$, $c$ in the following way.

$\dfrac {a \text{ } f(a^2 + 8bc) + b \text{ } f(b^2 + 8ca) + c \text{ } f(c^2 + 8ab)}{a+b+c} \geq f \left (\dfrac {a(a^2 + 8bc) + b(b^2 + 8ca) + c(c^2 + 8ab)}{a+b+c} \right )$

This may be rewritten as

$\dfrac{a}{\sqrt{a^2 + 8bc}} + \dfrac{b}{\sqrt{b^2 + 8ca}} + \dfrac{c}{\sqrt{c^2 + 8ab}} \geq \dfrac {(a+b+c)^{\frac {3}{2}}}{\sqrt {a^3 + b^3 + c^3 + 24abc}}.$

Thus, it suffices to prove

$(a+b+c)^3 \geq a^3 + b^3 + c^3 + 24abc$

But we can expand this and simplify to get

$a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + 2abc \geq 8abc$

$(a+b)(b+c)(c+a) \geq 8abc$

Which is true by AM-GM. Thus proven.

- 5 years, 3 months ago

Problem 15

Let $x_1, x_2, x_3, \ldots, x_m >1$ ,where $m \in \mathbb{N^+}$, be positive integers such that $x_j < x_{j+1}$ $(1 \leq j < m)$.

Prove that $\displaystyle \sum_{i=1}^m \frac{1}{x_i^3} < 1$

- 5 years, 3 months ago

Another interesting solution.

We got $x_i \geq i+1 \Leftrightarrow \frac{1}{x_i^3} \leq \frac{1}{(i+1)^3} < \frac{1}{(i+1)^2} < \frac{1}{i(i+1)} = \frac{1}{i} - \frac{1}{i+1} \Leftrightarrow \frac{1}{x_i^3} < \frac{1}{i} - \frac{1}{i+1}(*)$ Thus taking the sum for $i = 1$ until $i = m$: $\displaystyle \sum_{i=1}^m \frac{1}{x_i^3} \stackrel{(*)}{<} \sum_{i=1}^m \frac{1}{i} - \frac{1}{i+1} = 1 - \frac{1}{m+1} <1$ Since $\displaystyle \sum_{i=1}^m \frac{1}{i} - \frac{1}{i+1}$ is a telescoping series. Hence proved

- 5 years, 3 months ago

$\sum_{i=1}^{m} \dfrac{1}{x_i^3} < \int_{i=1}^{\infty}\dfrac{1}{x^3}\text{ d}x = \dfrac{1}{2} < 1$

- 5 years, 3 months ago

Problem 14

Show that for non-negative real $x$, $y$ and $z$,

$2x^3 (x^3 + 8y^3) + 2y^3 (y^3 + 8z^3) + 2z^3 (z^3 + 8x^3) \geq 9x^4 (y^2 + z^2) + 9y^4 (z^2 + x^2) + 9z^4 (x^2 + y^2)$

- 5 years, 3 months ago

SOLUTION OF PROBLEM 14 $LHS - RHS = \displaystyle \sum_{cyc} (x-y)^4(x^2 + 4xy + y^2) \geq 0$ Hence proved

- 5 years, 3 months ago

@Chris Galanis Can you please tell how you came up with that factorization?

- 5 years, 3 months ago

Actually when I solve inequalities this way I expect $LHS-RHS \geq 0$. Since $x, y, z$ are non negative I expect all the negative terms to form a square. In particular I tried to find pattern like $a^2 - 2ab +b^2$ (which is $(a-b)^2$) or $a^4 - 4a^3b + 6a^2b^2 -4ab^3 + b^4$ (which is $(a-b)^4$). Just that

- 5 years, 3 months ago

Nice!

- 5 years, 3 months ago

Problem 13

Given that $a,b,c$ are non-negative reals satisfying $a+b+c=1$, prove that $(1-a)(1-b)(1-c)\le \dfrac{7}{27}+abc$

- 5 years, 3 months ago

Solution to Problem 13

I expanded (once more) to get a solution.

Replace $(1-a), (1-b), (1-c)$ with $(b+c), (c+a), (a+b)$ in the in equation to get:

$(a+b)(b+c)(c+a) \leq \dfrac {7}{27} + abc$

Expanding and simplifying, we get

$a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + abc \leq \dfrac {7}{27}$

Removing the denominator, we get

$27a^2b + 27a^2c + 27b^2a + 27b^2c + 27c^2a + 27c^2b + 27abc \leq 7$

Homogenising the RHS, we get

$27a^2b + 27a^2c + 27b^2a + 27b^2c + 27c^2a + 27c^2b + 27abc \leq 7(a+b+c)^3$

$27a^2b + 27a^2c + 27b^2a + 27b^2c + 27c^2a + 27c^2b + 27abc \leq 7a^3 + 7b^3 + 7c^3 + 21a^2b + 21a^2c + 21b^2a + 21b^2c + 21c^2a + 21c^2b + 42abc$

$6a^2b + 6a^2c + 6b^2a + 6b^2c + 6c^2a + 6c^2b \leq 7a^3+ 7b^3 + 7c^3 + 15abc$

$a^2(b + c) + b^2(a + c) + c^2(a + b) \leq \dfrac {7}{6} a^3 + \dfrac {7}{6} b^3 + \dfrac {7}{6} c^3 + \dfrac {5}{2} abc$

Note that by Schur's, $a^2(b + c) + b^2(a + c) + c^2(a + b) \leq a^3 + b^3 + c^3 + 3abc$. We will prove the sharper inequality:

$a^3 + b^3 + c^3 + 3abc \leq \dfrac {7}{6} a^3 + \dfrac {7}{6} b^3 + \dfrac {7}{6} c^3 + \dfrac {5}{2} abc$

$6a^3 + 6b^3 + 6c^3 + 18abc \leq 7a^3 + 7b^3 + 7c^3 + 15abc$

$3abc \leq a^3 + b^3 + c^3$

Which is obviously true by AM-GM. Thus, proven.

- 5 years, 3 months ago

Note that this problem is actually equivalent to IMO 1984 #1.

Extension: can you find the best constants for the RHS?

- 5 years, 3 months ago

PROBLEM 12

If $a,b,c,d$ are positive reals such that $abcd=1$, prove that $\dfrac{1}{(1+a)(1+a^{2})}+\dfrac{1}{(1+b)(1+b^{2})}+\dfrac{1}{(1+c)(1+c^{2})}+\dfrac{1}{(1+d)(1+d^{2})} \ge 1$

- 5 years, 3 months ago

Problem 11 :

If $a,b,c$ are positive real numbers that satisfy $a+b+c=1$, prove that $\sqrt{a+bc}+ \sqrt{b+ac} + \sqrt{c+ba} \leq 2$

- 5 years, 3 months ago

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)=1$ or $ab+bc+ca \le \dfrac{1}{3}$ $\color{#D61F06}{\text{Since}} \quad \color{#D61F06}{{a^{2}+b^{2}+c^{2} \ge ab+bc+ca}}$