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.
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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
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 :
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 :
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
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
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
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, ${a_1}\geq{a_2}\geq{a_3}\geq\cdots\geq{a_n}$