Cho hai số thực a, b thỏa mãn ${{a}^{2}}+{{b}^{2}}>1$ và ${{\log...

Do ${{a}^{2}}+{{b}^{2}}>1$ nên từ ${{\log }_{{{a}^{2}}+{{b}^{2}}}}\left( a+b \right)\ge 1\Leftrightarrow {{\log }_{{{a}^{2}}+{{b}^{2}}}}\left( a+b \right)\ge {{\log }_{{{a}^{2}}+{{b}^{2}}}}\left( {{a}^{2}}+{{b}^{2}} \right)$
$\Leftrightarrow a+b\ge {{a}^{2}}+{{b}^{2}}>1$.
Suy ra: $\left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}>1 \\
& {{\left( a-\dfrac{1}{2} \right)}^{2}}+{{\left( b-\dfrac{1}{2} \right)}^{2}}\le \dfrac{1}{2} \\
\end{aligned} \right.$
Khi đó: $P=2a+4b-3=2\left( a-\dfrac{1}{2} \right)+4\left( b-\dfrac{1}{2} \right)\le \sqrt{\left( {{2}^{2}}+{{4}^{2}} \right).\left[ {{\left( a-\dfrac{1}{2} \right)}^{2}}+{{\left( b-\dfrac{1}{2} \right)}^{2}} \right]}$
$\le \sqrt{20.\left( \dfrac{1}{2} \right)}=\sqrt{10}$. (Áp dụng BĐT Bunhiacốpxki)
Dấu "=" xảy ra khi: $\left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}>1 \\
& \dfrac{a-\dfrac{1}{2}}{2}=\dfrac{b-\dfrac{1}{2}}{4}>0 \\
& {{\left( a-\dfrac{1}{2} \right)}^{2}}+{{\left( b-\dfrac{1}{2} \right)}^{2}}=\dfrac{1}{2} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{1}{2}+\dfrac{1}{\sqrt{10}} \\
& b=\dfrac{1}{2}+\dfrac{2}{\sqrt{10}} \\
\end{aligned} \right.$.
Vậy ${{P}_{\max }}=\sqrt{10}$ khi $\left\{ \begin{aligned}
& a=\dfrac{1}{2}+\dfrac{1}{\sqrt{10}} \\
& b=\dfrac{1}{2}+\dfrac{2}{\sqrt{10}} \\
\end{aligned} \right.$