Bài 5:
\(a,\dfrac{a^2+b^2}{2}\ge ab\\ \Leftrightarrow a^2+b^2\ge2ab\\ \Leftrightarrow a^2-2ab+b^2\ge0\\ \Leftrightarrow\left(a-b\right)^2\ge0\left(luôn.đúng\right)\)
Dấu "=" xảy ra\(\Leftrightarrow a=b\)
\(b,\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge2\\ \Leftrightarrow\dfrac{a^4+b^4}{a^2b^2}\ge2\\ \Leftrightarrow a^4+b^4\ge2a^2b^2\\ \Leftrightarrow a^4-2a^2b^2+b^4\ge0\\ \Leftrightarrow\left(a^2-b^2\right)^2\ge0\)
Dấu "=" xảy ra\(\Leftrightarrow a=b\)
Bài 6:
\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\\ \Rightarrow\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\\ \Rightarrow\left(x^2+5x-6\right)\left(x^2+5x+6\right)\\ \Rightarrow A=\left(x^2+5x\right)^2-36\ge-36\)
Dấu "=" xảy ra \(\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(A_{min}=-36\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Bài 5:
a: Ta có: \(\left(a-b\right)^2>=0\)
\(\Leftrightarrow a^2+b^2-2ab>=0\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}>=ab\)(đpcm)
b: \(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge2\sqrt{\dfrac{a^2}{b^2}\cdot\dfrac{b^2}{a^2}}=2\)
