Inequality Marathon

For discussing Olympiad Level Algebra (and Inequality) problems
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Anindya Biswas
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Inequality Marathon

Unread post by Anindya Biswas » Tue Jun 01, 2021 1:17 am

Inequalities are getting common in recent olympiads and they are coming like a surprise. But many of us have little practice solving them. In this time, an inequality marathon would be a source to practice and solve inequality problems.
Rules of posting solutions and problems :
  1. $L^AT_EX$ is a must.
  2. Add source to the problem if available.
  3. Use proper numbering in posting problems and solutions.
  4. Hide the source and the solution.
  5. After posting a problem, it's solution must be posted within $48$ hours.
  6. After proposing a problem, if it stays unsolved after $48$ hours, the proposer would provide the solution and post a new one and move on.
  7. If no new posts are posted within $72$ hours, one can post a new problem.
Last edited by Anindya Biswas on Tue Jun 01, 2021 1:26 am, edited 2 times in total.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Anindya Biswas
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Problem 00

Unread post by Anindya Biswas » Tue Jun 01, 2021 1:24 am

Let $a,b,c$ be positive real numbers satisfying $abc\geq1$
Prove that, \[\frac{a^2bc}{\sqrt{bc}+1}+\frac{b^2ca}{\sqrt{ca}+1}+\frac{c^2ab}{\sqrt{ab}+1}\geq\frac32\]
Source :
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

~Aurn0b~
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Re: Problem 00

Unread post by ~Aurn0b~ » Tue Jun 01, 2021 11:32 am

Anindya Biswas wrote:
Tue Jun 01, 2021 1:24 am
Let $a,b,c$ be positive real numbers satisfying $abc\geq1$
Prove that, \[\frac{a^2bc}{\sqrt{bc}+1}+\frac{b^2ca}{\sqrt{ca}+1}+\frac{c^2ab}{\sqrt{ab}+1}\geq\frac32\]
Source :
$\textbf{Solution 00}$
Let $a=x^2,b=y^2,c=z^2$.

$\sum \frac{a^2bc}{\sqrt{bc}+1}=abc \sum \frac{a}{\sqrt{bc}+1}$
So it suffices to prove that,$\sum \frac{a}{\sqrt{bc}+1}\geq\frac{3}{2}\Rightarrow \sum \frac{x^2}{yz+1}\geq \frac{3}{2}$
Now,
$xy+yz+zx\geq 3(x^2y^2z^2)^\frac{1}{3}\geq 3(abc)^\frac{1}{3}\geq 3$

$\Rightarrow 3(xy+yz+zx)\geq 9$

$\Rightarrow 9 \leq 3(xy+yz+zx)\leq xy+yz+zx+2(x^2+y^2+z^2)$

$\Rightarrow 9+3(xy+yz+zx) \leq 4(xy+yz+zx)+ 2(x^2+y^2+z^2)$

$\Rightarrow \frac{3}{2}\leq\frac{(x+y+z)^2}{xy+yz+zx+3}\leq \sum \frac{x^2}{yz+1}.\blacksquare$

~Aurn0b~
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Re: Inequality Marathon

Unread post by ~Aurn0b~ » Tue Jun 01, 2021 11:42 am

$\textbf{Problem 01}$

Let $ x,y,z\in \mathbb{R}_+$ . Prove that :
\[ \sqrt {x(y + 1)} + \sqrt {y(z + 1)} + \sqrt {z(x + 1)}\le \frac {3}{2}\sqrt {(x + 1)(y + 1)(z + 1)}\]

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Solution to Problem 01

Unread post by Anindya Biswas » Tue Jun 01, 2021 10:48 pm

~Aurn0b~ wrote:
Tue Jun 01, 2021 11:42 am
$\textbf{Problem 01}$

Let $ x,y,z\in \mathbb{R}_+$ . Prove that :
\[ \sqrt {x(y + 1)} + \sqrt {y(z + 1)} + \sqrt {z(x + 1)}\le \frac {3}{2}\sqrt {(x + 1)(y + 1)(z + 1)}\]
Let $a=\sqrt{\frac{x}{x+1}}, b=\sqrt{\frac{y}{y+1}}, c=\sqrt{\frac{z}{z+1}}$
$\therefore a,b,c\in(0,1)$
By RMS-AM inequality,
\begin{equation}
\begin{split}
\sqrt{1+a^2-b^2}+\sqrt{1+b^2-c^2}+\sqrt{1+c^2-a^2}&\leq3\cdot\sqrt{\frac{(1+a^2-b^2)+(1+b^2-c^2)+(1+c^2-a^2)}{3}}\\
&=3\\
\therefore \sqrt{1+a^2-b^2}+\sqrt{1+b^2-c^2}+\sqrt{1+c^2-a^2}\leq3\cdots\cdots\cdots(0)
\end{split}
\end{equation}
By Cauchy-Schwartz's inequality,
\begin{equation}
\begin{split}
&\left(a\sqrt{1-c^2}+b\sqrt{1-a^2}\right)^2\leq(a^2+1-a^2)(b^2+1-c^2)\\
&\Rightarrow a\sqrt{1-c^2}+b\sqrt{1-a^2}\leq\sqrt{1+b^2-c^2}\cdots\cdots\cdots(1)
\end{split}
\end{equation}
Similarly,
\begin{equation}
\begin{split}
b\sqrt{1-a^2}+c\sqrt{1-b^2}&\leq\sqrt{1+c^2-a^2}\cdots\cdots\cdots(2)\\
c\sqrt{1-b^2}+a\sqrt{1-c^2}&\leq\sqrt{1+a^2-b^2}\cdots\cdots\cdots(3)
\end{split}
\end{equation}
By adding $(1), (2)$ and $(3)$,
\begin{equation}
\begin{split}
2a\sqrt{1-c^2}+2b\sqrt{1-a^2}+2c\sqrt{1-b^2}&\leq\sqrt{1+a^2-b^2}+\sqrt{1+b^2-c^2}+\sqrt{1+c^2-a^2}\\
&\leq3\\
\Rightarrow 2\sqrt{\frac{x}{(x+1)(z+1)}}+2\sqrt{\frac{y}{(y+1)(x+1)}}+2\sqrt{\frac{z}{(z+1)(y+1)}}&\leq3\\
\Rightarrow \sqrt{x(y+1)}+\sqrt{y(z+1)}+\sqrt{z(x+1)}&\leq\frac32\sqrt{(x+1)(y+1)(z+1)}\blacksquare
\end{split}
\end{equation}
Last edited by Anindya Biswas on Tue Jun 01, 2021 11:29 pm, edited 1 time in total.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Problem 02

Unread post by Anindya Biswas » Tue Jun 01, 2021 10:55 pm

Reposting an old problem...
Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=5$
Find the minimum value of $\sqrt{a^2+1}+\sqrt{b^2+4}+\sqrt{c^2+16}+\sqrt{d^2+25}$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Re: Problem 02

Unread post by Anindya Biswas » Fri Jun 04, 2021 12:44 am

Anindya Biswas wrote:
Tue Jun 01, 2021 10:55 pm
Reposting an old problem...
Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=5$
Find the minimum value of $\sqrt{a^2+1}+\sqrt{b^2+4}+\sqrt{c^2+16}+\sqrt{d^2+25}$
Solution :
$\text{Lemma :}$ \[\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2}\geq\sqrt{\left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2}\] For all real numbers $a_1,b_1,a_2,b_2$ with equality if and only if $a_1b_2-a_2b_1=0$.
$\text{Proof :}$
Let's assume $a_1,b_1,a_2,b_2\in\mathbb{R}$
\[
\begin{equation}
\begin{split}
& (a_1b_1+a_2b_2)^2+(a_1b_2-a_2b_1)^2 \geq (a_1b_1+a_2b_2)^2 \\
& \Longrightarrow \left(a_1^2+b_1^2\right)\left(a_2^2+b_2^2\right)\geq (a_1b_1+a_2b_2)^2 \\
& \Longrightarrow \sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}\geq \lvert a_1b_1+a_2b_2\rvert \geq a_1b_1+a_2b_2 \\
& \Longrightarrow a_1^2+b_1^2+a_2^2+b_2^2+2\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)} \geq a_1^2+2a_1b_1+b_1^2+a_2^2+2a_2b_2+b_2^2 \\
& \Longrightarrow \left(\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2}\right)^2 \geq \left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2 \\
& \Longrightarrow \sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2} \geq \sqrt{\left(a_1+a_2\right)^2+\left(b_1+b_2\right)^2} \\ &\text{Q.E.D.}
\end{split}
\end{equation}
\]

According to this lemma,
\begin{equation}
\begin{split}
\sqrt{a^2+1}+\sqrt{b^2+4}+\sqrt{c^2+16}+\sqrt{d^2+25}&\geq\sqrt{(a+b+c+d)^2+(1+2+4+5)^2}\\
&=13.
\end{split}
\end{equation}
And this value is achievable when $a=\dfrac5{12}, b=\dfrac56, c=\dfrac53, d=\dfrac{25}{12}$.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Problem 03

Unread post by Anindya Biswas » Fri Jun 04, 2021 1:45 am

Let $a_1,a_2,a_3,\cdots,a_n$ be positive real numbers where $n\geq2, n\in\mathbb{N}$.
Let $s=a_1+a_2+a_3+\cdots+a_n$.
Prove that \[\frac{a_1}{s-a_1}+\frac{a_2}{s-a_2}+\frac{a_3}{s-a_3}+\cdots+\frac{a_n}{s-a_n}\geq\frac{n}{n-1}\]
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Mahiir
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Solution of Problem 03

Unread post by Mahiir » Thu Jul 01, 2021 8:20 pm

Anindya Biswas wrote:
Fri Jun 04, 2021 1:45 am
Let $a_1,a_2,a_3,\cdots,a_n$ be positive real numbers where $n\geq2, n\in\mathbb{N}$.
Let $s=a_1+a_2+a_3+\cdots+a_n$.
Prove that \[\frac{a_1}{s-a_1}+\frac{a_2}{s-a_2}+\frac{a_3}{s-a_3}+\cdots+\frac{a_n}{s-a_n}\geq\frac{n}{n-1}\]
$Solution$:

$WLOG$,

Let us consider, ${a_1}\geq{a_2}\geq{a_3}\geq\cdots\geq{a_n}$

So

${s-a_1}\leq{s-a_2}\leq{s-a_3}\leq\cdots\leq{s-a_n}$

$⇒\frac{1}{s-a_1}\geq\frac{1}{s-a_2}\geq\frac{1}{s-a_3}\geq\cdots\geq\frac{1}{s-a_n}$
Then...

$By$ $Rearrangement$ $Inequality,$

We can say that-


$\frac{a_1}{s-a_1}+\frac{a_2}{s-a_2}+\frac{a_3}{s-a_3}+\cdots+\frac{a_n}{s-a_n}$ is in it's highest value rearrangement.

Then by RI

$\frac{a_1}{s-a_1}+\frac{a_2}{s-a_2}+\frac{a_3}{s-a_3}+\cdots+\frac{a_n}{s-a_n}\geq\frac{a_2}{s-a_1}+\frac{a_3}{s-a_2}+\frac{a_4}{s-a_3}+\cdots+\frac{a_1}{s-a_n}$

$\frac{a_1}{s-a_1}+\frac{a_2}{s-a_2}+\frac{a_3}{s-a_3}+\cdots+\frac{a_n}{s-a_n}\geq\frac{a_3}{s-a_1}+\frac{a_4}{s-a_2}+\frac{a_5}{s-a_3}+\cdots+\frac{a_2}{s-a_n}$

$\frac{a_1}{s-a_1}+\frac{a_2}{s-a_2}+\frac{a_3}{s-a_3}+\cdots+\frac{a_n}{s-a_n}\geq\frac{a_4}{s-a_1}+\frac{a_5}{s-a_2}+\frac{a_6}{s-a_3}+\cdots+\frac{a_3}{s-a_n}$
$\vdots$
$\vdots$
$\frac{a_1}{s-a_1}+\frac{a_2}{s-a_2}+\frac{a_3}{s-a_3}+\cdots+\frac{a_n}{s-a_n}\geq\frac{a_n}{s-a_1}+\frac{a_1}{s-a_2}+\frac{a_2}{s-a_3}+\cdots+\frac{a_{n-1}}{s-a_n}$

Summing these all together , we get-

$(n-1)\times(\frac{a_1}{s-a_1}+\frac{a_2}{s-a_2}+\frac{a_3}{s-a_3}+\cdots+\frac{a_n}{s-a_n})\geq\frac{s-a_1}{s-a_1}+\frac{s-a_2}{s-a_2}+\frac{s-a_3}{s-a_3}+\cdots+\frac{s-a_{n}}{s-a_n}$

$⇒$ $ \frac{a_1}{s-a_1}+\frac{a_2}{s-a_2}+\frac{a_3}{s-a_3}+\cdots+\frac{a_n}{s-a_n}\geq\frac{n}{n-1}$

$Q.E.D$

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Mahiir
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Problem 04

Unread post by Mahiir » Thu Jul 01, 2021 8:34 pm

Let $a$,$b$,$c$ be positive real numbers such that $a+b+c=3$. Prove that

$a^2+b^2+c^2\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$

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