Áp dụng BĐT cô-si, ta có \(a^3+b^3+c^3\ge3abc\Rightarrow\frac{a^3+b^3+c^3}{2abc}\ge\frac{3}{2}\)
Mà \(\frac{a^2+b^2}{c^2+ab}\ge\frac{a^2+b^2}{c^2+\frac{a^2+b^2}{2}}=2\frac{a^2+b^2}{2c^2+a^2+b^2}\)
tương tự thì \(P\ge\frac{3}{2}+2\left(\frac{a^2+b^2}{2c^2+a^2+b^2}+\frac{b^2+c^2}{2a^2+b^2+c^2}+\frac{c^2+a^2}{2b^2+a^2+c^2}\right)\)
Đặt \(\hept{\begin{cases}a^2+b^2=x\\b^2+c^2=y\\c^2+a^2=z\end{cases}}\)
ta có \(P\ge\frac{3}{2}+2\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=\frac{3}{2}+2\left(\frac{x^2}{xy+xz}+\frac{y^2}{yz+yx}+\frac{z^2}{zx+zy}\right)\)
=>\(P\ge\frac{3}{2}+2.\frac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\frac{3}{2}+2.\frac{3\left(xy+yz+zx\right)}{2\left(xy+yz+zx\right)}\ge\frac{3}{2}+3=\frac{9}{2}\)
dấu xảy ra <>a=b=c>0
Vậy ...
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