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\(P=2\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)\ge\frac{2.4}{x^2+y^2+2xy}=\frac{8}{\left(x+y\right)^2}=8\)
Dấu "=" khi \(x=y=\frac{1}{2}\)
1) Biến đồi tương đương:
\(\left(x^2+y^2\right)^2\ge8\left(x-y\right)^2\)
\(\Leftrightarrow\left(x^2+y^2\right)^2\ge8xy\left(x-y\right)^2\)
\(\Leftrightarrow\left(x^2-4xy+y^2\right)^2\ge0\)(đúng)
2) Sửa đề: \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\left(\text{với }xy\ge1\right)\)
\(\Leftrightarrow\frac{\left(x-y\right)^2\left(xy-1\right)}{\left(x^2+1\right)\left(y^2+1\right)\left(xy+1\right)}\ge0\) (đúng)
Áp dụng bđt AM-GM ta có:
\(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\)
\(\Rightarrow\left(x+\frac{1}{x}\right)^2\ge4\)
CMTT \(\left(y+\frac{1}{y}\right)^2\ge4\)
\(\Rightarrow\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge4\left(dpcm\right)\)
Dấu"="xảy ra \(\Leftrightarrow x=y=1\)
Xét \(\frac{x}{y^3-1}+\frac{y}{x^3-1}=\frac{1-y}{y^3-1}+\frac{1-x}{x^3-1}=-\frac{1}{x^2+x+1}-\frac{1}{y^2+y+1}\)
\(=-\frac{x^2+y^2+x+y+2}{\left(x^2+x+1\right)\left(y^2+y+1\right)}=-\frac{x^2+y^2+3}{x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+x+y+1}\)
\(=-\frac{\left(x+y\right)^2-2xy+3}{x^2y^2+x^2+y^2+2xy+2}=-\frac{4-2xy}{x^2y^2+3}=\frac{2\left(xy-2\right)}{x^2y^2+3}\)
từ đó ta có đpcm
\(\dfrac{x-y}{z^2+1}=\dfrac{x-y}{z^2+xy+yz+zx}=\dfrac{x-y}{z\left(z+y\right)+x\left(z+y\right)}=\dfrac{x-y}{\left(x+z\right)\left(z+y\right)}\)
Tương tự: \(\dfrac{y-z}{x^2+1}=\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}\);\(\dfrac{z-x}{y^2+1}=\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
Cộng vế với vế \(\Rightarrow VT=\dfrac{x-y}{\left(x+z\right)\left(y+z\right)}+\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}+\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\dfrac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(đpcm)
Đặt: \(A=\frac{1}{x}+\frac{1}{y}+\frac{2}{x+y}\)
Ta có: \(A=\frac{1}{x}+\frac{1}{y}+\frac{2}{x+y}=\frac{xy}{x}+\frac{xy}{y}+\frac{2}{x+y}\left(\text{Do: xy = 1}\right)\)
\(=x+y+\frac{2}{x+y}\)
\(=\frac{x+y}{2}+\frac{x+y}{2}+\frac{2}{x+y}\)
Đặt: \(B=\frac{x+y}{2};C=\frac{x+y}{2}+\frac{2}{x+y}\)
\(\Rightarrow A=B+C\)
Vì x, y > 0, áp dụng BĐT Cô-si, ta có:
\(\Rightarrow B=\frac{x+y}{2}\ge\sqrt{xy}=\sqrt{1}=1\) (1)
Ta có: x, y > 0 => x + y > 0
Áp dụng BĐT \(\frac{a}{b}+\frac{b}{a}\ge2\) với hai số dương x + y và 2
\(\Rightarrow C=\frac{x+y}{2}+\frac{2}{x+y}\ge2\) (2)
\(\text{Từ (1); (2) }\Rightarrow B+C=\frac{x+y}{2}+\frac{2}{x+y}\ge1+2\)
\(\Rightarrow A\ge3\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{2}{x+y}\ge3\)
=> ĐPCM