Question

cho 3 so thuc duong  d e f thoa man \(\frac{d}{e+f}+\frac{e}{d+f}+\frac{f}{d+e}=\frac{3}{2}\)

CMR d=e=f

Answer 1

Áp dụng bất đẳng thức Nesbitt với 3 số dương d,e,f ta có: \(\frac{d}{e+f}+\frac{e}{d+f}+\frac{f}{d+e}\ge\frac{3}{2}\)

Dấu "=" xảy ra khi d=e=f

Answer 2

Chứng minh rằng \(\frac{d}{e+f}+\frac{e}{d+f}+\frac{f}{d+e}\ge\frac{3}{2}\)\(\forall d,e,f>0\)

\(\Rightarrow\frac{d}{e+f}+1+\frac{e}{d+f}+1+\frac{f}{d+e}+1\ge\frac{9}{2}\)

\(\Rightarrow\frac{d+e+f}{e+f}+\frac{d+e+f}{d+f}+\frac{d+e+f}{d+e}\ge\frac{9}{2}\)

\(\Rightarrow\left(d+e+f\right)\left(\frac{1}{e+f}+\frac{1}{d+f}+\frac{1}{d+e}\right)\ge\frac{9}{2}\)

\(\Rightarrow2\left(d+e+f\right)\left(\frac{1}{e+f}+\frac{1}{d+f}+\frac{1}{d+e}\right)\ge9\)

\(\Rightarrow\left(e+f+d+f+d+e\right)\left(\frac{1}{e+f}+\frac{1}{d+f}+\frac{1}{d+e}\right)\ge9\)

Áp dụng bất đẳng thức Cauchy 

\(\Rightarrow\left(e+f+d+f+d+e\right)\left(\frac{1}{e+f}+\frac{1}{d+f}+\frac{1}{d+e}\right)\ge9\sqrt[3]{\left(e+f\right)\left(d+f\right)\left(d+e\right).\frac{1}{\left(e+f\right)\left(d+f\right)\left(d+e\right)}}=9\)

Vậy ta có đpcm 

Dấu " = " xảy ra khi \(e=d=f\) ( đpcm )