Cho $x,y\in \left( 0;\dfrac{\pi }{2} \right)$ thỏa mãn $3{{\tan...

Ta có $3{{\tan }^{2}}x+2{{\cot }^{2}}y=5\left( \dfrac{60}{4\cos x+9\sin y}-7 \right)=\dfrac{300}{4\cos x+9\sin y}-35$
Áp dụng BĐT Cô-si với 5 số dương a, b, c, d, e ta có:
$\dfrac{1}{a+b+c+d+e}\le \dfrac{1}{5.\sqrt[5]{a.b.c.d.e}}=\dfrac{1}{5}.\sqrt[5]{\dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{c}.\dfrac{1}{d}.\dfrac{1}{e}}\le \dfrac{1}{25}.\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}+\dfrac{1}{e} \right)$
Vì $x,y\in \left( 0;\dfrac{\pi }{2} \right)$ nên $\dfrac{300}{4\cos x+9\sin y}=\dfrac{300}{2\cos x+2\cos x+3\sin y+3\sin y+3\sin y}$
$\Rightarrow \dfrac{300}{4\cos x+9\sin y}\le \dfrac{300}{25}\left( \dfrac{1}{2\cos x}+\dfrac{1}{2\cos x}+\dfrac{1}{3\sin y}+\dfrac{1}{3\sin y}+\dfrac{1}{3\sin y} \right)=12\left( \dfrac{1}{\cos x}+\dfrac{1}{\sin y} \right)$
Do đó $3{{\tan }^{2}}x+2{{\cot }^{2}}y\le 12\left( \dfrac{1}{\cos x}+\dfrac{1}{\sin y} \right)-35$
$\Leftrightarrow \dfrac{3}{{{\cos }^{2}}x}+\dfrac{2}{{{\sin }^{2}}y}-12\left( \dfrac{1}{\cos x}+\dfrac{1}{\sin y} \right)+30\le 0$
$\Leftrightarrow 3{{\left( \dfrac{1}{\cos x}-2 \right)}^{2}}+2{{\left( \dfrac{1}{\sin y}-3 \right)}^{2}}\le 0\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{\sin y}-3=0 \\
& \dfrac{1}{\cos x}-2=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \sin y=\dfrac{1}{3} \\
& \cos x=\dfrac{1}{2} \\
\end{aligned} \right.$
Vậy $P={{\cos }^{4}}x+{{\sin }^{4}}y=\dfrac{1}{16}+\dfrac{1}{81}=\dfrac{97}{1296}.$
$T=a+b=97+1296=1393.$