\(\dfrac{bc}{a^2+3}=\dfrac{bc}{a^2+b^2+a^2+c^2}\le\dfrac{1}{4}\left(\dfrac{\left(b+c\right)^2}{a^2+b^2+a^2+c^2}\right)\le\dfrac{1}{4}\left(\dfrac{b^2}{a^2+b^2}+\dfrac{c^2}{a^2+c^2}\right)\)
Tương tự ta có:
\(\dfrac{ca}{b^2+3}\le\dfrac{1}{4}\left(\dfrac{c^2}{b^2+c^2}+\dfrac{a^2}{a^2+b^2}\right)\)
\(\dfrac{ab}{c^2+3}\le\dfrac{1}{4}\left(\dfrac{a^2}{a^2+c^2}+\dfrac{b^2}{b^2+c^2}\right)\)
Cộng vế:
\(A\le\dfrac{1}{4}\left(\dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{a^2+b^2}+\dfrac{b^2}{b^2+c^2}+\dfrac{c^2}{b^2+c^2}+\dfrac{a^2}{a^2+c^2}+\dfrac{c^2}{a^2+c^2}\right)=\dfrac{3}{4}\)
\(A_{max}=\dfrac{3}{4}\) khi \(a=b=c=1\)
b.
\(B=\left(x-y^2\right)+x^2+4y^2+5\ge5\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\x^2=0\\y^2=0\end{matrix}\right.\) \(\Rightarrow x=y=0\)
