Prove that $\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$

Question

For $a,b,c$ are positive real number. Prove that

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

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Let $\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)$

We need prove $\sqrt{2x}+\sqrt{2y}+\sqrt{2z}\le\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$

We have: $\left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)^2=2\left(x+y+z\right)+2\left[\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(y+z\right)\left(z+x\right)}+\sqrt{\left(z+x\right)\left(x+y\right)}\right]$

By C-S we have: $\sqrt{\left(x+y\right)\left(y+z\right)}\ge\sqrt{xy}+\sqrt{yz}$

$\Rightarrow 2\sum\sqrt{\left(x+y\right)\left(y+z\right)}\ge4\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)$

$\Rightarrow LHS^2\ge 2(x+y+z+2\sqrt {xy}+2\sqrt {yz}+2\sqrt {xz})=2(\sqrt{x}+\sqrt{y}+\sqrt{z})=RHS^2$

Can do other way ?

Answer 1

we know that: \begin{align*} \left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2&\geq0\\ or,\ \frac{2}{\sqrt{ab}}&\leq\left(\frac{1}{a}+\frac{1}{b}\right)\\ or,\ \frac{1}{a}+\frac{1}{b}+\frac{2}{\sqrt{ab}}&\leq2\left(\frac{1}{a}+\frac{1}{b}\right)\\ or,\ \left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)^2&\leq2\left(\frac{1}{a}+\frac{1}{b}\right)\\ or,\ \frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}&\leq\sqrt{2\left(\frac{1}{a}+\frac{1}{b}\right)} \end{align*}

Similarly, $\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\leq\sqrt{2\left(\dfrac{1}{b}+\dfrac{1}{c}\right)}\ $ and, $\ \dfrac{1}{\sqrt{c}}+\dfrac{1}{\sqrt{a}}\leq\sqrt{2\left(\dfrac{1}{c}+\dfrac{1}{a}\right)}$

By adding them, \begin{align*} 2\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\right)&\leq \sqrt{2}\left[\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}+\sqrt{\left(\dfrac{1}{b}+\dfrac{1}{c}\right)}+\sqrt{\left(\dfrac{1}{c}+\dfrac{1}{a}\right)}\right]\\ or,\ \sqrt{\dfrac{2}{a}}+\sqrt{\dfrac{2}{b}}+\sqrt{\dfrac{2}{c}}&\leq\sqrt{\dfrac{a+b}{ab}}+\sqrt{\dfrac{b+c}{bc}}+\sqrt{\dfrac{c+a}{ca}} \end{align*}

Answer 2

Hint:

Note $t \to \sqrt t$ is concave and $\left( 2x, 2y, 2z \right) \succ \left( x+y, y+z, z+x \right) $, so Karamata's inequality applies.

Answer 3

Hint: prove the following first (e.g. by squaring), then add up all the combinations thereof:

$$ \sqrt{\dfrac{1}{a}}+\sqrt{\dfrac{1}{b}} \le \sqrt{2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)} $$

Answer 4

Apply the popular inequality $\sqrt{\dfrac{a^2+b^2}{2}}\ge \dfrac{a+b}{2}$, with $a = \sqrt{2x}, b = \sqrt{2y}$, and do the same for the other two pairs $(b,c)$ and $(c,a)$, and add all $3$ inequalities to get the result.