Câu 1:
a, Giả sử \(A=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-\dfrac{a}{b}-\dfrac{b}{a}\ge0\)
\(\Leftrightarrow A=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-2\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge0\)
Mà \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\Leftrightarrow A\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-2\cdot\dfrac{a}{b}-2\cdot\dfrac{b}{a}+2\ge0\)
\(\Leftrightarrow\left(\dfrac{a^2}{b^2}-2\cdot\dfrac{a}{b}+1\right)+\left(\dfrac{b^2}{a^2}-2\cdot\dfrac{b}{a}+1\right)\ge0\\ \Leftrightarrow\left(\dfrac{a}{b}-1\right)^2+\left(\dfrac{b}{a}-1\right)^2\ge0\left(\text{luôn đúng}\right)\)
Dấu \("="\Leftrightarrow a=b\)
b, \(B=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\right)+2+\left(\dfrac{a^2}{b^2}+2+\dfrac{b^2}{a^2}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)-4\)
\(B=\left(\dfrac{a^4}{b^4}-2\cdot\dfrac{a^2}{b^2}+1\right)+\left(\dfrac{b^4}{a^4}-2\cdot\dfrac{b^2}{a^2}+1\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)-2\\ \Leftrightarrow B=\left(\dfrac{a^2}{b^2}-1\right)^2+\left(\dfrac{b^2}{a^2}-1\right)^2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2+\dfrac{a}{b}+\dfrac{b}{a}-4\\ \Leftrightarrow B\ge0+0+0+\dfrac{a^2+b^2}{ab}-4\ge\dfrac{2ab}{ab}-4=2-4=-2\)
Dấu \("="\Leftrightarrow\left(a;b\right)\in\left\{\left(1;-1\right);\left(-1;1\right)\right\}\)
Câu 2:
\(\left(x^2+y^2\right)\left(3^2+4^2\right)\ge\left(3x+4y\right)^2=M^2\\ \Leftrightarrow M^2\le25\cdot25\\ \Leftrightarrow M\le25\)
Dấu \("="\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{25}{25}=1\Leftrightarrow\left\{{}\begin{matrix}x^2=9\\y^2=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
Vậy \(M_{max}=25\Leftrightarrow\left(x;y\right)=\left(3;4\right)\)
