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ข้อความเดิมเขียนโดยคุณ No.Name

3. $a,b,c >0$

$$\sqrt{\dfrac{a^3}{a^3+(b+c)^3}}+\sqrt{\dfrac{b^3}{b^3+(c+a)^3}}+\sqrt{\dfrac{c^3}{c^3+(a+b)^3}} \ge 1$$

let $x=\frac{a}{b+c},y=\frac{b}{c+a},z=\frac{c}{a+b}$
Then It's Equavalent to $$\sqrt{\frac{x^3}{x^3+1}}+\sqrt{\frac{y^3}{y^3+1}}+\sqrt{\frac{z^3}{z^3+1}}\ge 1$$
use $f(x)=\frac{x^3}{x^3+1}$ then $f$ is convex by Jense's Ineuality
$$\sqrt{\frac{x^3}{x^3+1}}+\sqrt{\frac{y^3}{y^3+1}}+\sqrt{\frac{z^3}{z^3+1}} \ge 3\sqrt{\frac{(\frac{x+y+z}{3})^3}{(\frac{x+y+z}{3})^3+1}}$$
Then It's Remain to show that $x+y+z\ge \frac{3}{2}$
That's Nesbitt's Ineqaulity