An easy inequality from Balkan MO SL 2016 A1

Let $a,b,c$ be positive real numbers. Prove that $\sqrt{a^3b+a^3c}+\sqrt{b^3c+b^3a}+\sqrt{c^3a+c^3b}\ge \frac43 (ab+bc+ca).$

#Algebra

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

Solution in the reply section:

ChengYiin Ong - 1 month, 1 week ago

By GM-HM inequality, we have $\sum_{cyc} \sqrt{a^2\cdot a(b+c)} \ge \sum_{cyc} \frac{2}{\frac{1}{a^2}+\frac{1}{a(b+c)}}=\sum_{cyc} \frac{2a^2(b+c)}{a+b+c}.$ The last sum is greater than or equal to $\frac{4}{3}(ab+bc+ca)$ ,

$\begin{aligned} &\sum_{cyc} \frac{2a^2(b+c)}{a+b+c} \ge \frac{4}{3}(ab+bc+ca) \\ \iff &a^2b+a^2c+b^2c+b^2a+c^2a+c^2b\ge 6abc \end{aligned}$

which is true by a simple AM-GM. $\quad \blacksquare$

ChengYiin Ong - 1 month, 1 week ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$