# Brilliant Inequality Contest - Season 2

Welcome all to Season 2 of the 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, but non-standard inequalities are also allowed, (due to increasing number of non-standard inequalities found in olympiads)

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. Most Standard Inequalities are allowed to be used, including AM-GM, Muirhead, Power Mean and Weighted Power Mean, Cauchy-Schwarz, Holder, Rearrangement, Chebyshev, Schur, Jensen, Karamata, Reverse Rearrangement, Bernoulli's Inequality 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. Also remember to keep the comments sorted by Newest, so you can see the current problem. And above all else, have fun!!!

PROBLEM 1

For real numbers $a_1, a_2, \ldots$, if $a_{n-1} + a_{n+1} \geq 2a_n$ for $n\geq 2$, then prove that

$A_{n-1} + A_{n+1} \geq 2A_n \quad \text{ for } n\geq 2.$

where $A_n$ is the average of $a_1, a_2, \ldots, a_n$. Note by Sharky Kesa
3 years ago

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Problem 4

Prove that for all non-negative real numbers $x,y,z$,

$6(x+y-z)(x^2+y^2+z^2)+27xyz \le 10(x^2+y^2+z^2)^{3/2}$

Applying Cauchy-Schwarz inequality, we have:

$\quad (x^2+y^2+z^2)(2^2+2^2+1^2)\ge(2x+2y+z)^2$

or $3(x^2+y^2+z^2)^{1/2}\ge 2x+2y+z$.

Hence,

$\quad10(x^2+y^2+z^2)^{3/2}-6(x+y-z)(x^2+y^2+z^2)$

$\ge(x^2+y^2+z^2)\left(\dfrac{10}{3}(2x+2y+z)-6(x+y-z)\right)$

$=\dfrac{1}{3}(x^2+y^2+z^2)(2x+2y+28z)$

Appying AM-GM inequality, we have:

$x^2+y^2+z^2=4\cdot\dfrac{x^2}{4}+4\cdot\dfrac{y^2}{4}+z^2\ge9\sqrt{\dfrac{x^8y^8z^2}{4^8}}$

$2x+2y+28z=2x+2y+7\cdot 4z\ge 9\sqrt{2x\cdot2y\cdot(4z)^7}=9\sqrt{4^8xyz^7}$

This implies that $(x^2+y^2+z^2)(2x+2y+28z)\ge 81xyz$, we are done.

The equality holds if and only if $x=y=2z$.

- 3 years ago

Problem 6:

Let $a$, $b$ and $c$ be positive reals such that $abc=1$. Prove that: $1+\sqrt{(a^3+1)(b^3+1)(c^3+1)}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$

- 3 years ago

Solution:

\begin{aligned} 1+\sqrt{(a^3+1)(b^3+1)(c^3+1)} &\geq \dfrac{a}{b} + \dfrac {b}{c} + \dfrac {c}{a}\\ (a^3+1)(b^3+1)(c^3+1) &\geq (ab+bc+ca-1)^3\\ \displaystyle \sum_{\text{cyc}} a^3 + 3 \displaystyle \sum_{\text{cyc}} a^2 b^2 + 6 \displaystyle \sum_{\text{cyc}} a^2 bc &\geq 3 \displaystyle \sum_{\text{sym}} a^2 b + 3 \displaystyle \sum_{\text{cyc}} ab + 3 \end{aligned}

The rest is just homogenisation and Muirhead bashing, which I'll update later since I have to go.

- 3 years ago

Can you update your complete solution?

- 3 years ago

let us change first n-1 -> let's say z then $a_{z}$ + $a_{z+2}$ >= 2 $a _{z+1}$ , now $A_{n}$ = $a_{1}$ ...........+ $a_{z}$ if we repeatedly add the given condition in the question i.e. $a_{z}$ + $a_{z+2}$ >= 2 $a _{z+1}$ you may refer to these pics but the handwriting is very poor :P! You have quite a few typos, but the essence of the proof is there, so I think you have solved it. I would recommend just going over the proof and fixing up the small errors you made to make it better. Good job. :)

- 3 years ago

For anyone interested in my proof of this problem, the algebra bash is fairly straightforward (clear out denominators, subtract and add, but be careful not to make errors). We manipulate $A_{n-1} + A_{n+1} \geq 2A_{n}$ to

$a_{1} + a_{2} + \ldots + a_{n-1} - \dfrac {n^2+n-2}{2} a_n + \dfrac {n(n-1)}{2} a_{n+1} \geq 0$

However, we can convert this into the telescoping sum

$=\displaystyle \sum_{k=2}^n \dfrac{k(k-1)}{2}(a_{k-1} - 2a_{k} + a_{k+1}) \geq 0$

The final expression is true by our original condition, so we are done.

- 3 years ago

yes, good one !

$\large 2^{nd} ques$  if a,b,c are the sides of a triangle provided with perimeter of this triangle i.e 10  $\large Prove\ that :$  $\large { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+\frac { 2 }{ 5 } abc\quad <\quad 50$

Solution to Problem 2

Let $a=x+y$, $b=y+z$, $c=z+x$. Since $a$, $b$ and $c$ are sides of a triangle, $x$, $y$ and $z$ must be positive by incentre properties. Also, since $a+b+c=10$, $x+y+z=5$. Substituiting this in to the inequation we wish to prove, we have:

\begin{aligned} (x+y)^2 + (y+z)^2 + (z+x)^2 + \dfrac{2}{5} (x+y)(y+z)(z+x) &< 50\\ 5x^2 + 5y^2 + 5z^2 + 5xy + 5yz + 5zx + x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y + 2xyz &< 125\\ (x+y+z)(x^2 + y^2 + z^2 + xy + yz + zx) + x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y + 2xyz &< (x+y+z)^3\\ x^3 + y^3 + z^3 + 3(x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y) + 5xyz &< x^3 + y^3 + z^3 + 3(x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y) + 6xyz\\ 0 &< xyz\\ \end{aligned}

We have already said that $x$, $y$ and $z$ are positive, so the final expression is obviously true. Therefore, the first expression must be true. Thus, proven.

For further details, what I have done is use the incircle substitution so I can get rid of the weird triangle condition. Since every triangle has an incircle, and tangents to a circle from a point are equal, we can use this substitution.

In the second half of my proof, I homogenised all the terms by substituting $x+y+z=5$ so they were all degree 3, and they simplified down to the final expression.

- 3 years ago

Yeah Ravi substitution is best when dealing with triangle sides inequalities !

- 3 years ago

Problem 3

Let $a$, $b$ and $c$ be positive reals such that $abc \leq 1$. Prove that

$\dfrac {a}{c} + \dfrac {b}{a} + \dfrac {c}{b} \geq a + b + c$

- 3 years ago

By AM-GM inequality,

$\dfrac{a}{b}+\dfrac{c}{a}+\dfrac{c}{a} \ge 3 \sqrt{\frac{c^2}{ab}} \ge 3 \sqrt{c^3} = 3c$

Similarly, $\dfrac{b}{c}+\dfrac{a}{b}+\dfrac{a}{b} \ge 3 \sqrt{\frac{a^2}{bc}} \ge 3 \sqrt{a^3} = 3a$

and $\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{b}{c} \ge 3 \sqrt{\frac{b^2}{ca}} \ge 3 \sqrt{b^3} = 3b$

Adding all the equaions, $3 \left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right) \ge 3(a+b+c)$.

Hence, $\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \ge a+b+c$ with equality iff $a=b=c=1$.

Problem 5:

Let $a,b,c$ be real numbers such that $a+b+c=0$.

Prove that:

$a^2b^2+b^2c^2+c^2a^2+6abc+3\ge0$

- 3 years ago

48 hours have passed since I posted this problem. Here is a solution.

WLOG, assume that $ab\ge 0$.

From $a+b+c=0$, we have $c=-(a+b)$.

Substituiting this in to the inequation we wish to prove, we have:

$\quad (a^2+b^2)(a+b)^2+a^2b^2-6ab(a+b)+3\ge0$

or $(a^2+ab+b^2)^2-6ab(a+b)+3\ge 0$.

Appying AM-GM inequality, we have:

$\quad (a^2+ab+b^2)^2-6ab(a+b)+3$

$\ge \dfrac{(a^2+ab+b^2)^2}{3}+\dfrac{2(a^2+ab+b^2)^2}{3}-\dfrac{3ab[(a+b)^2+4]}{2}+3$

$\ge \dfrac{9a^2b^2}{3}+2ab(a^2+ab+b^2)-\dfrac{3ab(a+b)^2}{2}-6ab+3$

$=3(ab-1)^2+\dfrac{ab(a-b)^2}{2}\ge0$.

We have equality for $(a,b,c)=(1,1,-2)$ and its permutations.

- 3 years ago

Problem 7:

Given distinct positive integers $a_1, a_2, \ldots, a_n$, prove that

$(a_1^7 + a_2^7 + \ldots + a_n^7) + (a_1^5 + a_2^5 + \ldots + a_n^5) \geq 2(a_1^3 + a_2^3 + \ldots + a_n^3)^2$

Also, find the equality case.

- 3 years ago

Solution:

We try to prove this via induction.

For $n=1$, we have

$a_1^7 + a_1^5 - 2(a_1^3)^2 = a_1^5 (a_1-1)^2 \geq 0$

so the base case $n=1$ is true. Suppose the case is true for $n=k$. For the case $n=k+1$, we first WLOG $a_1 < a_2 < \ldots < a_{k+1}$. We have

\begin{aligned} 2(a_1^3+a_2^3+\ldots + a_{k+1}^3)^2 - 2(a_1^3+a_2^3+\ldots + a_k^3)^2 &= 2a_{k+1}^6 + 4a_{k+1}^3 (a_1^3 + \ldots + a_k^3)\\ & \leq 2a_{k+1}^6 + 4a_{k+1}^3 (1^3 + 2^3 + \ldots + (a_{k+1}-1)^3)\\ &= 2a_{k+1}^6 + 4a_{k+1}^3 \dfrac{(a_{k+1}-1)^2 a_{k+1}^2}{4}\\ &= a_{k+1}^7 + a_{k+1}^5\\ \end{aligned}

Thus, $(a_1^7 + a_2^7 + \ldots + a_n^7) + (a_1^5 + a_2^5 + \ldots + a_n^5) \geq 2(a_1^3 + a_2^3 + \ldots + a_n^3)^2$, with equality iff $a_1 =1, a_2 = 2, \ldots, a_n = n$.

- 3 years ago

Problem 8:

Let $a_1 \geq a_2 \geq \ldots \geq a_n \geq a_{n+1} = 0$ be a sequence of real numbers. Prove that

$\sqrt \sum_{k=1}^n a_{k}} \leq \displaystyle \sum_{k=1}^n \sqrt{k} (\sqrt{a_k} - \sqrt{a_{k+1}}$

- 3 years ago

Solution:

Let $x_k = \sqrt{a_k} - \sqrt{a_{k+1}}$. Then, $a_k = (x_k + x_{k+1} + \ldots + x_{n})^2$. Thus,

\begin{aligned} \displaystyle \sum_{k=1}^n &= \displaystyle \sum_{k=1}^n (x_k + x_{k+1} + \ldots + x_n)^2 = \displaystyle \sum_{k=1}^n kx_k^2 + 2 \displaystyle \sum_{1 \leq i \leq j \leq n} i x_i x_j\\ &\leq \displaystyle \sum_{k=1}^n kx_k^2 + 2\displaystyle \sum_{1\leq i \leq j \leq n} \sqrt{ij}x_i x_j = \left ( \displaystyle \sum_{k=1}^n \sqrt{k} x_k \right ) ^2\\ \end{aligned}

Taking the square root of both sides gives us the desired inequality.

- 3 years ago

Problem 9:

For $0 \leq a \leq b \leq c \leq d \leq e$ and $a+b+c+d+e=1$, show that

$ad + dc + cb + be +ea \leq \dfrac {1}{5}$

- 3 years ago

Solution:

Since, $a\le b\le c\le d\le e$, we get $a+b\le a+c\le b+d\le c+e\le d+e$.

Applying the Chebyshev's inequality, we have:

$\quad2(a+b+c+d+e)^2$

$=\left(a+b+c+d+e\right)\left((d+e)+(c+e)+(b+d)+(a+c)+(a+b)\right)$

$\ge 5(ad+ae+bc+be+cb+cd+da+dc+ea+eb)$

$=10(ad+dc+cb+be+ea)$

Hence, $ad+dc+cb+be+ea\le\dfrac{1}{5}(a+b+c+d+e)^2=\dfrac{1}{5}$.

The equality holds when $a=b=c=d=e=\dfrac{1}{5}$.

- 3 years ago

Problem 10:

Let $a,b,c,d,e$ be non-negative real numbers such that $a+b+c+d+e=5$. Prove that: $abc+bcd+cde+dea+eab\le5$

- 3 years ago

Solution:

Assume $e=\min\{a, b, c, d,e\}$. Then AM-GM gives :

$e(c + a)(b + d) + bc(a + d - e) \le \dfrac{e(5 - e)^2}{4}+\dfrac{(5 - 2e)^2}{27}\le 5$

the last one being equivalent with: $(e - 1)^2(e + 8)\ge 0$, which is true.

- 3 years ago

Problem 11:

Let $a,b$ and $c$ be positive reals such that $a+b+c=1$. Prove that: $\sqrt{a+\dfrac{(b-c)^2}{4}}+\sqrt b+\sqrt c\le\sqrt3$

- 3 years ago

- 3 years ago

Applying the Cauchy-Schwarz inequality, we have:

$\quad\left(\sqrt{a+\dfrac{(b-c)^2}{4}}+\sqrt b+\sqrt c\right)^2$

$\le 3\left[a+\dfrac{(b-c)^2}{4}+\dfrac{\left(\sqrt b+\sqrt c\right)^2}{4}+\dfrac{\left(\sqrt b+\sqrt c\right)^2}{4}\right]$

$=3\left[a+\dfrac{(b-c)^2+2(b+c)+4\sqrt{bc}}{4}\right]$

$=3\left[a+\dfrac{2(b+c)+\left(\sqrt b+\sqrt c\right)^2\left(\sqrt b-\sqrt c\right)^2+4\sqrt{bc}}{4}\right]$

$\le 3\left[a+\dfrac{2(b+c)+2(b+c)\left(\sqrt b-\sqrt c\right)^2+4\sqrt{bc}}{4}\right]$

$\le 3\left[a+\dfrac{2(b+c)+2\left(\sqrt b-\sqrt c\right)^2+4\sqrt{bc}}{4}\right]$

$\le 3\left[a+\dfrac{2(b+c)+2(b+c)}{4}\right]=3$, Q.E.D.

The equality occurs iff $a=b=c=\dfrac{1}{3}$.

- 3 years ago

Post the next problem as well.

- 3 years ago

Problem 12:

Let $a,b$ and $c$ be real numbers such that: $-1\le a,b,c\le 1$ and $a+b+c=0$.

Prove that: $a^2+b^4+c^6\le 2$.

- 3 years ago

I'm not sure whether this hould keep running, due to lack of participants. Thoughts?

- 2 years, 12 months ago

Sorry for late reply. In a recent days, my family and I are welcoming the Vietnamese New Year.

This is the solution:

Because $a^2+b^4+c^6\le a^2+b^2+c^2$, we can prove a stronger inequality: $a^2+b^2+c^2\le 2$.

WLOG, assume that $a\le b\le c$. hence $-1\le a\le 0\le c\le 1$.

We have: $a(a+1)\le 0$ or $a^2\le -a$; $c(c-1)\le 0$ or $c^2\le c$.

If $-1\le b\le 0$, we get $b^2\le -b$, so $a^2+b^2+c^2\le -a-b+c=2c\le 2$.

If $0\le b\le 1$, we get $b^2\le b$, so $a^2+b^2+c^2\le -a+b+c=-2a\le 2$.

Equility holds when $(a,b,c)=(-1,0,1)$ and its permutations.

- 2 years, 12 months ago

Sharky, is the rules number 9 are all the inequality formula that have been discovered ?

I'm just here to try problem and study XD

Even though i didn't solve one of them

- 2 years, 12 months ago

There are so many more inequalities. Those listed are just the standard dozen and some extra.

- 2 years, 12 months ago