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trinhminh18

Ta có: $a+(b+c+d)$ \geq $2\sqrt{a}.\sqrt{b+c+d}$ (theo cauchy)
\Rightarrow $\sqrt{a}(a+b+c+d)$ \geq $2a\sqrt{b+c+d}$
\Rightarrow $\dfrac{\sqrt{a}}{\sqrt{b+c+d}}$ \geq $\dfrac{2a}{a+b+c+d}$
c/m tương tự có
$\dfrac{\sqrt{b}}{\sqrt{a+c+d}}$\geq $\dfrac{2b}{a+b+c+d}$
$\dfrac{\sqrt{c}}{\sqrt{b+a+d}}$\geq $\dfrac{2c}{a+b+c+d}$
$\dfrac{\sqrt{d}}{\sqrt{b+c+a}}$\geq $\dfrac{2d}{a+b+c+d}$
\Rightarrow $\dfrac{\sqrt{a}}{\sqrt{b+c+d}}+ \dfrac{\sqrt{b}}{\sqrt{a+c+d}}+ \dfrac{\sqrt{c}}{\sqrt{b+a+d}} + \dfrac{\sqrt{d}}{\sqrt{b+c+a}}$ \geq $\dfrac{2a}{a+b+c+d}+ \dfrac{2b}{a+b+c+d} +\dfrac{2c}{a+b+c+d}+\dfrac{2d}{a+b+c+d} $
\Rightarrow đpcm
 
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