Chứng minh rằng: \(\frac{\sqrt{a+b+c}+\sqrt{a}}{b+c}+\frac{\sqrt{a+b+c}

Ta có: ĐPCM
\(\Leftrightarrow \frac{a+b+c+\sqrt{a(a+b+c)}}{b+c}+\frac{a+b+c+\sqrt{b(a+b+c)}}{c+a}+\frac{a+b+c+\sqrt{c(a+b+c)}}{a+b}\)\(\geq \frac{9+3\sqrt{3}}{2}\)
\(\Leftrightarrow \frac{1+\sqrt{\frac{a}{a+b+c}}}{\frac{b}{a+b+c}+\frac{c}{a+b+c}}+\frac{1+\sqrt{\frac{b}{a+b+c}}}{\frac{c}{a+b+c}+\frac{a}{a+b+c}}+\frac{1+\sqrt{\frac{c}{a+b+c}}}{\frac{a}{a+b+c}+\frac{b}{a+b+c}}\geq \frac{9+3\sqrt{3}}{2}\)
Đặt \(x=\frac{a}{a+b+c}; y=\frac{b}{a+b+c}; z=\frac{c}{a+b+c}\) ta có: x, y, z > 0 và x + y + z =1
Khi đó
ddpcm \(\Leftrightarrow \frac{1+\sqrt{x}}{y+z}+\frac{1+\sqrt{y}}{z+x}+\frac{1+\sqrt{z}}{x+y}\geq \frac{9+3\sqrt{3}}{2}\)
\(\Leftrightarrow \frac{1+\sqrt{x}}{1-x}+\frac{1+\sqrt{y}}{1-y}+\frac{1+\sqrt{z}}{1-z}\geq \frac{9+3\sqrt{3}}{2}\)
Ta cm: \(\Leftrightarrow \frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z}\geq \frac{9}{2}\) (1) Ta có:

\(\left [ (1-x)+(1-y)+(1-z) \right ]\left [ \frac{1}{1-x}+\frac{1}{1-y} +\frac{1}{1-z}\right ]\geq 9\)
\(\Rightarrow \frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z} \geq \frac{9}{2}\)
Từ đó suy ra (1) đúng, dấu đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)
Ta cm: \(\frac{\sqrt{x}}{1-x}+\frac{\sqrt{y}}{1-y}+\frac{\sqrt{z}}{1-z}\geq \frac{3\sqrt{3}}{2} \ \ (2)\)
Thật vậy, Xét hàm số \(f(x)=\sqrt{x}(1-x)\) với 0 < x < 1
Ta có \(f'(x)=\frac{1-3x}{2\sqrt{x}}=0\Leftrightarrow x=\frac{1}{3}\)
BBT

Suy ra \(0<f(x)<\frac{2}{3\sqrt{3}}\). Dấu “=” xảy ra \(\Leftrightarrow x=\frac{1}{3}\)
Vậy ta có \(\frac{\sqrt{x}}{1-x}=\frac{x}{(1-x)\sqrt{x}}\geq \frac{x}{2}=\frac{3\sqrt{3}x}{2}\)
tương tự \(\frac{\sqrt{y}}{1-y}\geq \frac{3\sqrt{3}x}{2}; \frac{\sqrt{z}}{1-z}\geq \frac{3\sqrt{3}z}{2}\)
Suy ra \(\frac{\sqrt{x}}{1-x}+\frac{\sqrt{y}}{1-y}+\frac{\sqrt{z}}{1-z}\geq \frac{3\sqrt{3}}{2}(x+y+z)=\frac{3\sqrt{3}}{2}\)
Từ đó suy ra (2) đúng, dấu đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)
Từ đó suy ra đpcm dấu đẳng thức xảy ra khi a=b=c