https://tranvantoancv.violet.vn/present/show/entry_id/10776977
Với x,y,z >0, ta có:
-\(\frac{x}{y}+\frac{y}{x}\ge2\) (1)
-\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\) (2)
-\(x^2+y^2+z^2\ge xy+yz+zx\Leftrightarrow\frac{x^2+y^2+z^2}{xy+yz+zx}\ge1\)(3)
Xảy ra đẳng thức ở (1),(2),(3) \(\Leftrightarrow x=y=z\), ta có:
\(P=\frac{ab+bc+ca}{a^2+b^2+c^2}+\left(a+b+c\right)^2.\frac{a+b+c}{abc}\)
=\(\frac{ab+bc+ca}{a^2+b^2+c^2}+\left(a^2+b^2+c^2+2ab+2bc+2ca\right).\frac{\left(a+b+c\right)}{abc}\)
Áp dụng các bất đẳng thức (1),(2),(3). ta có:
\(P\)\(\ge\)\(\frac{ab+bc+ca}{a^2+b^2+c^2}+\left(a^2+b^2+c^2\right).\frac{9}{ab+bc+ca}+2.9\)
=\(\left(\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{a^2+b^2+c^2}{ab+bc+ca}\right)+8.\frac{a^2+b^2+c^2}{ab+bc+ca}+18\)\(\ge2+8+18=28\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a^2+b^2+c^2=ab+bc+ca\\ab=bc=ca\end{cases}\Leftrightarrow a=b=c}\)
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