# 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
3 years, 7 months ago

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Solution of Problem 1 :

We will show that the given inequality is true for all positive reals $x,y,z$ satisfying the conditions given in the question.

Case 1: When x,y,z are sides of triangle.

Let $\color{#3D99F6}{x=a+b,y=b+c,z=a+c}$; $\text{(a,b,c>0)}$,so our inequality becomes:

$\color{#69047E}{\sum_{cyc}2(a+b)^2(a+b+2c) \geq \sum_{cyc} (a+b)^3 +9(a+b)(b+c)(a+c)}$

$\implies 2 \sum_{cyc}(a+b)^{3} + 4\sum_{cyc} c(a+b)^{2} \geq \sum_{cyc}(a+b)^3 +9(a+b)(b+c)(a+c)$

$\implies \sum_{cyc} (a+b)^{3} +4\sum_{cyc} c(a+b)^{2} \geq 9(a+b)(b+c)(a+c)$

$\implies 2 \sum_{cyc}a^{3}+ 3\sum_{cyc} (ab^{2} +ba^{2}) + 4\sum_{cyc} (ab^{2} +ba^{2}) + 24abc \geq 9(a+b)(b+c)(a+c)$

$\implies 2 \sum_{cyc} a^{3} + 7(a+b)(b+c)(c+a) + 10abc \geq 9(a+b)(b+c)(c+a)$

$\implies 2 \sum_{cyc} a^{3}+ 10abc \geq 2(a+b)(b+c)(c+a)$

$\implies \sum_{cyc} a^{3}+ 5abc \geq (a+b)(b+c)(c+a)$

$\color{#D61F06}{\implies \sum_{cyc} a^{3} + 3abc \geq a^{2} b + b^{2} a + a^{2} c + c^{2} a + b^{2} c + c^{2} b}$

which is true by Schur's Inequality.

Equality occurs when $a=b=c$ , that is $x=y=z$.

Hence proved.

Case 2: When x,y,z are not sides of a triangle.

In this case either $x+y=z$ or $y+z = x$ or $z+x=y$.

Let's take $\color{#3D99F6}{x+y =z}$.

Since the inequality is symmetric, WLOG we can assume that $\color{#20A900}{z \ge x \ge y}$.

Now substituting $x+y = z$ in the inequality , we get $\color{#D61F06}{ 2x^{2}(2y+x) + 2y^{2}(2x+y) + 2(x+y)^{3} \geq x^{3} + y^{3} + (x+y)^{3} + 9xyz}$

After rearranging the expression, we get,$x^{3} + y^{3} \geq x^{2} y + y^{2}x$

which is true by Rearrangement Inequality

Hence proved.

- 3 years, 7 months ago

Can you write the equality cases for Case 2?

- 3 years, 7 months ago

I don't think equality exists . . .

- 3 years, 7 months ago

Problem 8

For positive reals $a,b,c$, prove that $\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\left(\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\le \dfrac{1}{abc}$

- 3 years, 7 months ago

Solution to Problem 8

I used a massive expansion, but I've triple-checked so its right (If you haven't figured it out, expansion is what I do with inequalities). We have to prove

$\left (\dfrac {1}{a+b} + \dfrac{1}{b+c} \right ) \left (\dfrac {1}{b+c} + \dfrac{1}{c+a} \right ) \left (\dfrac {1}{c+a} + \dfrac{1}{a+b} \right ) \leq \dfrac {1}{abc}$

$\left (\dfrac {a+2b+c}{(a+b)(b+c)} \right ) \left (\dfrac {b+2c+a}{(b+c)(c+a)} \right ) \left (\dfrac {c+2a+b}{(c+a)(a+b)} \right ) \leq \dfrac {1}{abc}$

Simplifying the denominators, we have

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

Simplifying all the brackets, we get

\begin{aligned} &2a^4bc + 2ab^4c + 2abc^4 + 7a^3b^2c + 7a^3bc^2 + 7a^2b^3c + 7ab^3c^2 + 7a^2bc^3 + 7ab^2c^3 + 16a^2b^2c^2 \leq\\ &2a^4bc + 2ab^4c + 2abc^4 + 6a^3b^2c + 6a^3bc^2 + 6a^2b^3c + 6ab^3c^2 + 6a^2bc^3 + 6ab^2c^3 + 10a^2b^2c^2 \\& \quad \quad \quad \quad \quad+a^4b^2 + a^4c^2 + b^4a^2 + b^4c^2 + c^4a^2 + c^4b^2 + 2a^3b^3 + 2b^3c^3 + 2c^3a^3 \end{aligned}

Simplifying, we get

$a^3b^2c + a^3bc^2 + a^2b^3c + ab^3c^2 + a^2bc^3 + ab^2c^3 + 6a^2b^2c^2 \leq a^4b^2 + a^4c^2 + b^4a^2 + b^4c^2 + c^4a^2 + c^4b^2+2a^3b^3+2b^3c^3+2c^3a^3$

By AM - GM, we have

$\dfrac {a^4c^2 + a^2b^4}{2} \geq a^3b^2c, \quad \dfrac {a^4b^2 + a^2c^4}{2} \geq a^3bc^2, \quad \dfrac {b^4c^2 + b^2a^4}{2} \geq a^2b^3c$

$\dfrac {b^4a^2 + b^2c^4}{2} \geq ab^3c^2, \quad \dfrac {c^4b^2 + c^2a^4}{2} \geq a^2bc^3, \quad \dfrac {c^4a^2 + c^2b^4}{2} \geq ab^2c^3$

$a^3b^3 + b^3c^3 + c^3a^3 \geq 3a^2b^2c^2 \quad \Rightarrow \quad 2a^3b^3 + 2b^3c^3 + 2c^3a^3 \geq 6a^2b^2c^2$

Summing these, we get the expansion above. Thus, the statement is true.

- 3 years, 7 months ago

Consider this:

We just need to prove $\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\left(\dfrac{c}{b+c}+\dfrac{c}{c+a}\right)\left(\dfrac{a}{c+a}+\dfrac{a}{a+b}\right)\le 1$ However by AM-GM $LHS\le \left(\begin{array}{c}\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{c}{c+a}+\dfrac{a}{c+a}+\dfrac{a}{a+b}\\ \hline 3\end{array}\right)^3=1$ A beautiful solution, not by me :P

- 3 years, 7 months ago

Problem 3
Prove the following inequality if $a,b,c$ are positive reals and $a+b+c+abc=4$ ,
$(1+\dfrac{a}{b} +ac)(1+\dfrac{b}{c} +ba)(1+\dfrac{c}{a} +cb) \geq 27$

###### Pakistan Round 1-2016

- 3 years, 7 months ago

Problem 6 :

Let $x > y > z$ be positive reals greater than one.

Prove that $\large{ \sum_{cyc} x^{\frac{7}{3}} > \sum_{cyc} x^{2} y^{\frac{1}{3}}}$

- 3 years, 7 months ago

Solution 6: Thank You @Harsh Shrivastava,

Since we are given $x,y,z \in \mathbb{R^{+} }$ and $x>y>z$.

We have two sets $x^2 > y^2 > z^2$ and $x^{\frac{1}{3}} > y^{\frac{1}{3}} > z^{\frac{1}{3}}$.

By rearrangement inequality we have

$\Large{{ x }^{ 2 }{ x }^{ \frac { 1 }{ 3 } }+{ y }^{ 2 }{ y }^{ \frac { 1 }{ 3 } }+{ z }^{ 3 }{ z }^{ \frac { 1 }{ 3 } }>{ x }^{ 2 }{ y }^{ \frac { 1 }{ 3 } }+{ y }^{ 2 }{ z }^{ \frac { 1 }{ 3 } }+{ z }^{ 2 }{ x }^{ \frac { 1 }{ 3 } }\\ \therefore \quad \sum _{ \text{cyclic } }^{ }{ { x }^{ \frac { 7 }{ 3 } } } >\sum _{ \text{ cyclic } }^{ }{ { x }^{ 2 } } { y }^{ \frac { 1 }{ 3 } }}$

Now my time to post questions!! Harsh By the way how did you thought of this?

- 3 years, 7 months ago

Problem 13

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

- 3 years, 7 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.

- 3 years, 7 months ago

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

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

- 3 years, 7 months ago

Problem 5

Let $x$, $y$, $z$ be positive reals such that $xyz=1$. Show that

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

- 3 years, 7 months ago

Solution 5 :

$(x-y)^{2} \geq 0$

$\implies x^{2} + y^{2} \geq 2xy$

Adding $2x+2$ on both sides, $x^{2} + y^{2} +2x + 2 \geq 2xy +2x +2$

$\implies \dfrac{1}{xy+x+1} \geq \dfrac{2}{x^{2} + y^{2} +2x + 2 }$

$\implies \dfrac{1}{xy+x+1} \geq \dfrac{2}{(x+1)^2+y^2+1 }$

Similarly , $\dfrac{1}{yz+y+1} \geq \dfrac{2}{(y+1)^2+z^2+1 }$ and $\dfrac{1}{zx+z+1} \geq \dfrac{2}{(z+1)^2+x^2+1 }$

Adding the three inequalities, we get $\dfrac {2}{(x+1)^2+y^2+1} + \dfrac {2}{(y+1)^2+z^2+1} + \dfrac {2}{(z+1)^2+x^2+1} \leq \dfrac{1}{yz+y+1} + \dfrac{1}{xy+x+1}+ \dfrac{1}{xz+z+1}$

Since $\dfrac{1}{yz+y+1} + \dfrac{1}{xy+x+1}+ \dfrac{1}{xz+z+1} = 1$, the result is evident.

- 3 years, 7 months ago

Problem 14

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

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

- 3 years, 7 months ago

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

- 3 years, 7 months ago

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

- 3 years, 7 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

- 3 years, 7 months ago

Nice!

- 3 years, 7 months ago

Problem 25

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

- 3 years, 7 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.

- 3 years, 7 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 ;)

- 3 years, 7 months ago

PROBLEM 2 :

Let $a,b$ and $c$ be non-negative reals such that $a+b+c = 1$.

Show that $1 + 6abc \geq \dfrac{1}{4} + 3(a+b)(b+c)(c+a)$

- 3 years, 7 months ago

Homogenizing and expanding gives $a^3+b^3+c^3+6abc \ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2$ which is just Schur.

- 3 years, 7 months ago

Solution Problem 2:
$1+6abc \geq \dfrac{1}{4} +3(1-a)(1-b)(1-c)$
$1+6abc \geq \dfrac{1}{4} +3(1-(a+b+c)+\sum_{cyc} ab -abc)$
$1+6abc \geq \dfrac{1}{4} +3(\sum_{cyc} ab -abc)$
$\dfrac{3}{4}+9abc \geq 3 \times \sum_{cyc} ab$
$\dfrac{1}{4}+3abc - \sum_{cyc} ab \geq 0$
$1+ 12abc \geq 4 \sum_{cyc} ab$
$(a+b+c)^3+12abc \geq \sum_{cyc} ab$
$a^3+b^3+c^3+6abc \geq \sum_{sym} a^2b$
This is schurs inequality.

- 3 years, 7 months ago

How did you use $ab+bc+ca \le \dfrac{1}{3}$ on $1+12abc\ge 4(ab+bc+ca)$ to get $abc\ge 0$?

- 3 years, 7 months ago

Some of you might have seen this but still answer me.

Problem 7

Let $a, b, c$ be positive real numbers such that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c$. Prove that

$\Large{\frac { 1 }{ { \left( 2a+b+c \right) }^{ 2 } } +\frac { 1 }{ { \left( 2b+c+a \right) }^{ 2 } } +\frac { 1 }{ { \left( 2c+a+b \right) }^{ 2 } } \le \frac { 3 }{ 16 } }$

- 3 years, 7 months ago

Solution to Problem 7

By AM-GM $\dfrac{1}{(2a+b+c)^2} \le \dfrac{1}{4(a+b)(a+c)}$ and thus it remains to prove $\sum_{cyc} \dfrac{1}{(a+b)(a+c)}\le \dfrac{3}{4}$ Clearing denominators, it remains to prove $8(a+b+c)\le 3(a+b)(b+c)(c+a)$ and after homogenizing this becomes $8abc(a+b+c)^2 \le 3(ab+bc+ca)(a+b)(b+c)(c+a)$ which after expansion and simplification is $3\sum_{sym}a^3b^2 \ge \sum_{sym}a^3bc+2\sum_{sym}a^2b^2c$ which is true by Muirhead.

- 3 years, 7 months ago

Problem 10

Given that $a_2, a_3, \ldots , a_n$ is a permutation of $\{2, 3, \ldots , n\}$, prove that $(2a_2-1)(3a_3-1)\cdots (na_n-1)\ge \dfrac{(n-1)!(n+1)!}{2}$

- 3 years, 7 months ago

Solution 10 :

Observe that$\dfrac{(n-1)!(n+1)!}{2} = (n^{2}-1)((n-1)^{2} - 1) \cdots (3^{2}-1)(2^{2}-1)$

So now the inequality becomes $(2a_2-1)(3a_3-1)\cdots (na_n-1)\ge (n^{2}-1)((n-1)^{2} - 1) \cdots (3^{2}-1)(2^{2}-1)$

This can be proven by applying Reverse Rearrangement Inequality on the sets ${X} : (a_{2}-1),(a_{3}-1) \cdots, (a_{n}-1)$ and ${Y} : (1 \times a_{2}), (2\times a_{3}),(3\times a_{4}) \cdots,( (n-1) \times a_{n})$.

$X,Y$ will be similarly ordered when $a_{k} = k$.

So applying Reverse Rearrangement Inequality (Random Sum Product > Similar ordered Product), $\displaystyle\prod_{i=2}^n (X_{i}+Y_{i}) \geq \displaystyle\prod_{i=2}^n(i^{2} - 1)$

- 3 years, 7 months ago

PROBLEM 12

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

- 3 years, 7 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 $\displaystyle \sum_{i=1}^m \frac{1}{x_i^3} < 1$

- 3 years, 7 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

- 3 years, 7 months ago

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

- 3 years, 7 months ago

@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.

- 3 years, 6 months ago

Sure!

- 3 years, 6 months ago

More than welcome :)

- 3 years, 6 months ago

Waiting forit to begin...

- 3 years, 6 months ago

Problem 11 :

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

- 3 years, 7 months ago

SOLUTION OF PROBLEM 11

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

Now, by Cauchy-Schwarz inequality $\sqrt{\left((\sqrt{a+bc})^{2}+(\sqrt{b+ac})^{2}+(\sqrt{c+ba})^{2} \right)(1+1+1)} \ge \sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ba}$ or $\sqrt{3+3(ab+bc+ca)} \ge \sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ba}$

Now $\sqrt{3+3(ab+bc+ca)} \le 2$ since $ab+bc+ca \le \dfrac{1}{3}$. Hence $\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ba} \le 2$.

Equality holds when $a=b=c=\dfrac{1}{3}$. Hence proved.

- 3 years, 7 months ago

Can you please elaborate more on how you applied Cauchy-Schwarz here?

- 3 years, 7 months ago

It's a just applying Cauchy-Schwarz with $a+bc,b+ac,c+ba$ and (1,1,1) then, taking square root on both sides.

- 3 years, 7 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

- 3 years, 7 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

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