Theo Svac - xơ có :

\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge\frac{9}{ab+bc+ca}\)

Khi đó \(P\ge\frac{9}{ab+bc+ca}+\frac{1}{a^2+b^2+c^2}\)

\(=\left(\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{1}{a^2+b^2+c^2}\right)+\frac{7}{ab+bc+ca}\)

\(\ge\frac{9}{a^2+b^2+c^2+2.\left(ab+bc+ca\right)}+\frac{7}{\frac{\left(a+b+c\right)^2}{3}}\)

\(=\frac{9}{\left(a+b+c\right)^2}+\frac{21}{\left(a+b+c\right)^2}=\frac{30}{\left(a+b+c\right)^2}=\frac{10}{3}\)

Dấu "=: xảy ra khi \(a=b=c=1\)

Vậy \(P_{min}=\frac{10}{3}\) khi \(a=b=c=1\)

Bài 1: Ta có \(\left(\frac{a^2}{b}-a+b\right)+b^2=\frac{a^2-ab+b^2}{b}+b\ge2\sqrt{a^2-ab+b^2}\)  (áp dụng Bất Đẳng Thức Cosi)

\(=\sqrt{a^2-ab+b^2}+\sqrt{\frac{3}{4}\left(a-b\right)^2+\frac{1}{4}\left(a+b\right)^2}\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\)

\(\Rightarrow\frac{a^2}{b}-a+2b\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\left(1\right)\)

Tương tự ta có \(\hept{\begin{cases}\frac{b^2}{c}-b+2c\ge\sqrt{b^2-bc+c^2}+\frac{1}{2}\left(b+c\right)\left(2\right)\\\frac{c^2}{a}-c+2a\ge\sqrt{c^2-ac+a^2}+\frac{1}{2}\left(a+c\right)\left(3\right)\end{cases}}\)

Từ (1) và (2) và (3) \(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ac+a^2}\)

Dấu "=" xảy ra khi a=b=c

Dat \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x,y,z\right)\)

thi \(P= \Sigma \frac{z^2}{x+y} \geq \frac{x+y+z}{2} \) (1)

Mat khac co \(x+y+z=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=3\) (2)

Tu (1) va (2) suy ra \(P\ge\frac{3}{2}\).Dau = xay ra khi \(a=b=c=1\)

\(P=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a^2+b^2+c^2}\)

\(P=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{9-2\left(ab+bc+ca\right)}\)

\(P=\frac{1}{3ab}+\frac{1}{3bc}+\frac{1}{3ca}+\frac{1}{9-2\left(ab+bc+ca\right)}+\frac{2}{3}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(P\ge\frac{16}{3ab+3bc+3ca+9-2\left(ab+bc+ca\right)}+\frac{2}{3}\left(\frac{9}{ab+bc+ca}\right)\)

\(P\ge\frac{16}{9+ab+bc+ca}+\frac{6}{ab+bc+ca}\)

Sử dụng đánh giá quen thuộc:\(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\)

\(\Rightarrow ab+bc+ca\le3\)

\(\Rightarrow P\ge\frac{16}{9+3}+\frac{6}{3}=2+\frac{4}{3}=\frac{10}{3}\)

"="<=>a=b=c=1