Tìm $Min$ $D= \frac{x}{\sqrt{y+z -4}} +\frac{y}{\sqrt{z+x -4}}+ \frac{z}{\sqrt{x+y -4}}$

Question

Cho $x,y,z$ là những số thực $>2.$

Tìm $Min$

$D= \frac{x}{\sqrt{y+z -4}} +\frac{y}{\sqrt{z+x -4}}+ \frac{z}{\sqrt{x+y -4}}$

score 7 · Answer 1

Ta có:

$D=\sum_{}^{} \frac{2x}{\sqrt{4(y+z-4)}}\geq \sum_{}^{} \frac{2x}{\frac{4+y+x-4}{2}}=\sum_{}^{} \frac{4x}{y+z}$

$\geq 4.\sum_{}^{} \frac{x}{y+z}=4.\sum_{}^{}\frac{x^2}{xy+zx} \geq 4.\frac{(x+y+z)^2}{2(xy+yz+zx)}\geq 4.\frac{3}{2}=6.$

Vậy $Min$ $D=6\Leftrightarrow x=y=z=4$.

Bài toán xong !!!

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