Đặt \(\hept{\begin{cases}2x=a\left(a>0\right)\\3y=b\left(b>0\right)\end{cases}}\)
\(\Rightarrow2x+3y=a+b\le2,x.y=\frac{ab}{6}\)
\(\Rightarrow P=\frac{4}{a^2+b^2}+\frac{9}{\frac{ab}{6}}=\frac{4}{a^2+b^2}\ne\frac{54}{ab}\)
Vì \(a>0,b>0\)
Nên áp dụng BĐT cô-si ta có:\(a+b\ge2\sqrt{ab}\)
Mà \(a+b\le2\Rightarrow2\sqrt{ab}\le2\Rightarrow\sqrt{ab}\le1\Rightarrow ab\le1\)
Áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)với x > 0 , y > 0
\(\Rightarrow\frac{1}{a^2+b^2}+\frac{1}{2ab}\ge\frac{4}{a^2+b^2+2ab}=\frac{4}{\left(a+b\right)^2}\ge1\)
\(\Rightarrow\frac{4}{a^2+b^2}+\frac{4}{2ab}\ge4\)
\(\Rightarrow P=\frac{4}{a^2+b^2}+\frac{4}{2ab}+\frac{52}{ab}\)
\(P\ge4+52=56\)
\(\Rightarrow MinP=56\Leftrightarrow\hept{\begin{cases}a=b\\a+b=2\\a.b=1\end{cases}}\Leftrightarrow\hept{a=b=1\Leftrightarrow2x=3y=1\Leftrightarrow x=\frac{1}{2},y=\frac{1}{3}}\)