How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$

Solution 1:

Let $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$, where $x$, $y$ and $z$ are positives.

Hence, we need to prove that $$\sum_{cyc}\frac{\frac{x}{y}}{\sqrt{1+\frac{x}{y}}}\geq\frac{3}{\sqrt2}$$ or $$\sum_{cyc}\frac{x}{\sqrt{y(x+y)}}\geq\frac{3}{\sqrt2}.$$ Now, by Holder $$\left(\sum_{cyc}\frac{x}{\sqrt{y(x+y)}}\right)^2\sum_{cyc}xy(x+y)(2z+x+y)^3\geq\left(\sum_{cyc}x(2z+x+y)\right)^3.$$ Thus, it remains to prove that $$2\left(\sum_{cyc}(x^2+3xy)\right)^3\geq9\sum_{cyc}xy(x+y)(2z+x+y)^3$$ or $$\sum_{sym}(x^6+9x^5y+24x^4y^2+18x^3y^3+9x^4yz-36x^3y^2z-25x^2y^2z^2)\geq0,$$ which is obviously true.

Done!

Solution 2:

Consider the function for positive $x$: $$f(x) = \frac{x}{\sqrt{1+x}}-\frac1{\sqrt 2}-\frac3{4\sqrt 2}\log x$$

Note that $f(x) \ge 0 \implies f(a)+f(b)+f(c) \ge 0 \implies $ the given inequality. Now $$f'(x) = \frac{4x^2-3\sqrt2 (x+1)^{3/2}+8x}{8x(x+1)^{3/2}}$$

We need to check the sign of the numerator, $4(x+1)^2-3\sqrt2(x+1)^{3/2}-4$. Using $y = \sqrt{x+1}$, we get the numerator as

$$4y^4-3\sqrt2y^3-4 = (y-\sqrt2)(4y^3+\sqrt2y^2+2y+2\sqrt2)$$

As the second factor is positive, the numerator's sign is given by $y-\sqrt2$ which has the same sign as $x-1$, so $f'(x)< 0$ for $x < 1$ and $f'(x)> 0$ for $x> 1$. Hence $f(x)\ge f(1)=0$.

Solution 3:

By the same way as Mr. Mike$,$ it's enough to prove $$2\left(\sum_{cyc}(x^2+3yz)\right)^3\geqslant 9\sum xy(x+y)(2z+x+y)^3$$ Or $$\frac18\sum \left( 16\,{x}^{4}+100\,{z}^{4}+104\,{x}^{3}y+243\,{y}^{2}{z}^{2 }+330\,{z}^{3}x+416\,{y}^{2}zx+342\,{z}^{2}xy \right) \left( x-y \right) ^{2}+$$ $$+\frac18\sum x{y}^{2} \left( 18\,y+41\,x \right) \left( z+x-2\,y \right) ^{2}\geqslant 0$$