Không mất tính tổng quát gỉa sử:$a>b>c\geq0$Khi đó:$P\geq \frac{1}{(a-b)^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}$Khi đó:$3P= P.(\Sigma a^{2})\geq ( a^{2}+b^{2})\left[ {\frac{1}{(a-b)^{2}}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\right]=\frac{\frac{a}{b}+\frac{b}{a}}{\frac{a}{b}-2+\frac{b}{a}}+(\frac{a}{b}+\frac{b}{a})^{2}$Đặt $t=\frac{a}{b}+\frac{b}{a},(t>2)$$\Rightarrow P\geq f(t)=\frac{1}{3}(t^{2}+\frac{t}{t-2})$Xét h/s $f(t),t>2$$f'(t)=\frac{1}{3}\left[2t-\frac{2}{(t-2)^{2}} \right]$$f'(t)=0\Leftrightarrow t=\frac{3+\sqrt{5}}{2}$$Min P=\frac{11+\sqrt{5}}{6}\Leftrightarrow c=0;\frac{a}{b}+\frac{b}{a}=\frac{3+\sqrt{5}}{2}$Bn tự giải dấu ''='' xra nhé:Dmk hơi lười

Không mất tính tổng quát gỉa sử:$a>b>c\geq0$Khi đó:$P\geq \frac{1}{(a-b)^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}$Khi đó:$3P= P.(\Sigma a^{2})\geq ( a^{2}+b^{2})\left[ {\frac{1}{(a-b)^{2}}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\right]=\frac{\frac{a}{b}+\frac{b}{a}}{\frac{a}{b}-2+\frac{b}{a}}+(\frac{a}{b}+\frac{b}{a})^{2}$Đặt $t=\frac{a}{b}+\frac{b}{a},(t>2)$$\Rightarrow P\geq f(t)=\frac{1}{3}(t^{2}+\frac{t}{t-2})$Xét h/s $f(t),t>2$$f'(t)=\frac{1}{3}\left[2t-\frac{2}{(t-2)^{2}} \right]$$f'(t)=0\Leftrightarrow t=\frac{3+\sqrt{5}}{2}$$Min P=\frac{11+\sqrt{5}}{6}\Leftrightarrow c=0;\frac{a}{b}+\frac{b}{a}=\frac{3+\sqrt{5}}{2}$Dấu''='' xra$\Leftrightarrow c=0;a=\frac{3+\sqrt{5}\pm \sqrt{6\sqrt{5}-2}}{4}b$