Let's save this for Christmas!

Here's a new problem for you to think about, celebrating Chirstmas (although it has nothing to do with Christmas!).

$a$; $b$; $c$ are three positive real numbers such that $a + b + c = 1$. Prove that:

$\frac{\sqrt{abc}(\sqrt{a + b} + \sqrt{b + c} + \sqrt{c + a})}{ab + bc + ca} \le \sqrt 2$

#Algebra

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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}`	$\boxed{123}$

Comments

I actually combined $a^2 + b^2 + c^2 + \sqrt{12abc} \le 1$ and $\sqrt{1 - a} + \sqrt{1 - b} + \sqrt{1 - c} \le \sqrt 6$ .

So if there are any shorter solutions, it would be great.

Thành Đạt Lê - 3 years, 6 months 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}`	$\boxed{123}$