Chứng minh BĐT

S

soicon_boy_9x

Ta có:

$a+b+c+d \geq 2 \sqrt{a(b+c+d}$

$\leftrightarrow \dfrac{a}{\sqrt{a(b+c+d)}} \geq \dfrac{a}
{\dfrac{a+b+c+d}{2}}=\dfrac{2a}{a+b+c+d}$

Mà $ \dfrac{a}{\sqrt{a(b+c+d)}}=\sqrt{\dfrac{a}{b+c+d}}$

$\rightarrow \sqrt{\dfrac{a}{b+c+d}} \geq \dfrac{2a}{a+b+c+d}(1) $

Tương tự: $\sqrt{\dfrac{b}{a+c+d}} \geq \dfrac{2b}{a+b+c+d}(2) \\ \sqrt{\dfrac{c}{a+b+d}} \geq \dfrac{2c}{a+b+c+d}(3) \\ \sqrt{\dfrac{d}{a+b+c}} \geq \dfrac{2d}{a+b+c+d}(4)$

Cộng từng vế các bất đẳng thức cùng dấu $(1);(2);(3);(4)$ ta được dpcm
 
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