x,y,z>0; x+y+z=3
Tìm min
$\frac{x+1}{y^2+1}+\frac{y+1}{z^2+1}+\frac{z+1}{x^2+1}$
x,y,z>0; x+y+z=3
Tìm min
$\frac{x+1}{y^2+1}+\frac{y+1}{z^2+1}+\frac{z+1}{x^2+1}$
Ta có: $\frac{x+1}{y^2+1}=(x+1)-\frac{y^2(x+1)}{y^2+1}\geq x+1-\frac{y(x+1)}{2}$
Suy ra $\sum \frac{x+1}{y^2+1}\geq \frac{x+y+z-(xy+yz+zx)}{2}+3\geq 3$
$\sqrt[LOVE]{MATH}$
"If I feel unhappy, I do mathematics to become happy. If I am happy, I
do mathematics to keep happy" - Alfréd Rényi
$x+1+y+1+z+1-\frac{y^2(x+1)}{y^2+1}-\frac{z^2(y+1)}{z^2+1}-\frac{x^2(z+1)}{x^2+1}\geq 6-\frac{y(x+1)}{2}-\frac{z(y+1)}{2}-\frac{x(z+1)}{2}\geq 6-3=3$
Duyên do trời làm vương vấn một đời.
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