Question

1. cho x, y không âm thoả mãn X^2+ Y^2 = 1. tìm GTNN: A=\(\sqrt{4+5x}\) + \(\sqrt{4+5y}\)

2. với a, b không âm thoả mãn a^2 + b^2=4 . Tìm GTLN B= \(\frac{ab}{a+b+2}\)

 

Answer 1

\(B=\frac{ab}{a+b+2}\Rightarrow2B=\frac{2ab}{a+b+2}=\frac{\left(a+b\right)^2-a^2-b^2}{a+b+2}=\frac{\left(a+b\right)^2-4}{a+b+2}=a+b-2\)

Do a ; b không âm , áp dụng BĐT Cô - si cho 2 số , ta có :

\(a+b\le\sqrt{2\left(a^2+b^2\right)}=\sqrt{2.4}=\sqrt{8}\)

\(\Rightarrow a+b-2\le\sqrt{8}-2\)

\(\Rightarrow2B\le\sqrt{8}-2\Rightarrow B\le\sqrt{2}-1\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=\sqrt{2}\)

Answer 2

Do x ; y không âm , \(x^2+y^2=1\)

\(\Rightarrow\left|x\right|;\left|y\right|\le1\) \(\Rightarrow0\le x;y\le1\)

\(\Rightarrow x\ge x^2;y\ge y^2\Rightarrow x+y\ge x^2+y^2=1\)

\(x,y\ge0\Rightarrow xy\ge0\)

Ta có : \(A=\sqrt{5x+4}+\sqrt{5y+4}\)

\(\Rightarrow A^2=5x+4+5y+4+2\sqrt{\left(5x+4\right)\left(5y+4\right)}\)

\(=5\left(x+y\right)+8+2\sqrt{25xy+20y+20x+16}\)

\(\ge5.1+8+2\sqrt{25.0+20.1+16}=13+2.6=25\)

\(\Rightarrow A\ge5\)

Dấu " = " xảy ra \(\Leftrightarrow\left[{}\begin{matrix}x=0;y=1\\x=1;y=0\end{matrix}\right.\)