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Ta có: \(M=\left(\sqrt{a}+\sqrt{b}\right)^2\le\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2\)=\(2a+2b\le2\)
\(Max\)\(M=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}+\sqrt{b}\\a+b=1\end{matrix}\right.\)\(\Leftrightarrow a=b=\dfrac{1}{2}\)
\(M=\left(\sqrt[]{a}+\sqrt[]{b}\right)^2;a+b\le1\left(a;b>0\right)\)
Áp dụng Bất đẳng thức Bunhiacopxki cho 2 cặp số \(\left(1;\sqrt[]{a}\right);\left(1;\sqrt[]{b}\right)\)
\(M=\left(1.\sqrt[]{a}+1.\sqrt[]{b}\right)^2\le\left(1^2+1^2\right)\left(a+b\right)\le2\) \(\left(a+b\le1\right)\)
\(\Rightarrow M=\left(\sqrt[]{a}+\sqrt[]{b}\right)^2\le2\)
Dấu "=" xảy ra khi và chỉ khi
\(\dfrac{1}{\sqrt[]{a}}=\dfrac{1}{\sqrt[]{b}}\Leftrightarrow a=b=1\)
\(\Rightarrow GTLN\left(M\right)=2\left(khi.a=b=1\right)\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(P=\sqrt{\dfrac{yz}{x^2+1}}+\sqrt{\dfrac{zx}{y^2+1}}+\sqrt{\dfrac{xy}{z^2+1}}\)
\(P=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}+\sqrt{\dfrac{zx}{y^2+xy+yz+zx}}+\sqrt{\dfrac{xy}{z^2+xy+yz+zx}}\)
\(P=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\dfrac{zx}{\left(y+z\right)\left(x+y\right)}}+\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\)
\(P\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{z}{x+z}\right)+\dfrac{1}{2}\left(\dfrac{z}{y+z}+\dfrac{x}{x+y}\right)+\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)=\dfrac{3}{2}\)
\(P_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(a=b=c=\sqrt{3}\)