Prove this inequality $2(a+b+c)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$

Question

For $a,b,c$ are positive real numbers satisfy $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that $$2\left(a+b+c\right)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$$

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We have:$a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}\Leftrightarrow \:\left(a+b+c\right)^2\ge \:9\Leftrightarrow \:a+b+c\ge \:3$

By Cauchy-Schwarz: $R.H.S^2\le \left(1+1+1\right)\left(a^2+b^2+c^2+27\right)$

$=3\left(a^2+b^2+c^2+9\right)\Rightarrow R.H.S\le \sqrt{3\left(a^2+b^2+c^2\right)+27}$

Need to prove $4(a+b+2)^2\ge 3(a^2+b^2+c^2)+27$

$\Leftrightarrow \left(a+b+c\right)^2+3\left(a+b+c\right)^2\ge 3\left(a^2+b^2+c^2\right)+27$

$\Leftrightarrow \left(a+b+c\right)^2\ge 3\left(a^2+b^2+c^2\right)$ It's wrong. Help me

Answer 1

Since the function $f(t) := \sqrt{1+t}$ is concave in $[-1, +\infty)$, we have that $$ f(t) \leq f(3) + f'(3) (t-3) \qquad \forall t\geq -1, $$ i.e. $$ f(t) \leq 2 + \frac{1}{4}(t-3) = \frac{5}{4} + \frac{1}{4} t \qquad \forall t\geq -1. $$ Using this inequality we have that $$ \sqrt{a^2+3} = a \sqrt{1+ 3/a^2} \leq a \left[\frac{5}{4} + \frac{1}{4}\cdot\frac{3}{a^2}\right] = \frac{5}{4} a + \frac{3}{4}\cdot \frac{1}{a}, $$ and a similar inequality holds also for $\sqrt{b^2+3}$ and $\sqrt{c^2+3}$.

Finally, using the condition $a+b+c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$, $$ \sqrt{a^2+3} + \sqrt{b^2+3} + \sqrt{c^2+3} \leq \frac{5}{4} (a + b + c) + \frac{3}{4}\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = 2(a+b+c). $$

Answer 2

An alternative proof to that presented by @Rigel and one that does not require any results from calculus/differentiation is as follows:

Let $\mathbb{u}\equiv\left(\sqrt{a},\sqrt{b},\sqrt{c}\right)$ and $\mathbb{v}\equiv\left(\sqrt{a+\frac{3}{a}},\sqrt{b+\frac{3}{b}},\sqrt{c+\frac{3}{c}}\right)$. From these definitions one has: $$ |\mathbb{u}\cdot\mathbb{v}|=\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3},\qquad \mbox{and} $$ $$ ||\mathbb{u}||\!\cdot\!||\mathbb{v}||=\sqrt{a+b+c}\sqrt{a+b+c+3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}=\sqrt{a+b+c}\sqrt{4\left(a+b+c\right)}=2(a+b+c). $$ The result follows by application of the Cauchy-Schwarz Inequality.

Using the same method of proof, one can obtain the following more general inequality: $$ \sqrt{K+1}(a+b+c)\geq\sqrt{a^2+K}+\sqrt{b^2+K}+\sqrt{c^2+K}, $$ for any positive $K$.

Answer 3

We need to prove that $$\sum_{cyc}\left(2a-\sqrt{a^2+3}\right)\geq0$$ or $$\sum_{cyc}\frac{a^2-1}{2a+\sqrt{a^2+3}}\geq0$$ or $$\sum_{cyc}\left(\frac{a^2-1}{2a+\sqrt{a^2+3}}+\frac{1}{4}\left(\frac{1}{a}-a\right)\right)\geq0$$ or $$\sum_{cyc}(a^2-1)\left(\frac{1}{2a+\sqrt{a^2+3}}-\frac{1}{4a}\right)\geq0$$ 0r $$\sum_{cyc}\frac{(a^2-1)(2a-\sqrt{a^2+3})}{a(2a+\sqrt{a^2+3})}\geq0$$ or $$\sum_{cyc}\frac{(a^2-1)^2}{a(2a+\sqrt{a^2+3})^2}\geq0.$$ Done!