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Xét vế 1 ta có: \(\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{y}\) \(=\frac{yz+yx}{xz}+\frac{z+x}{y}\)
\(=\frac{y^2z+y^2x+x^2z+xz^2}{xyz}\)nhóm hạng tử 1 với 4,2 với 3 trên tử ta được:
\(=\frac{z\left(y^2+xz\right)+x\left(y^2+xz\right)}{xyz}\)\(=\frac{\left(z+x\right)\left(y^2+xz\right)}{xyz}=\frac{z+x}{zx}\times\frac{y^2+xz}{y}\)(1);
Xét vế 2 ta có: \(=1+\frac{x}{z}+\frac{z}{x}+1=2+\frac{x}{z}+\frac{z}{x}\)nhân 2 đa thức với nhau:
\(=\frac{2xz}{xz}+\frac{x^2+z^2}{xz}\)\(=\frac{x^2+2xz+z^2}{xz}\)\(=\frac{\left(x+z\right)^2}{xz}=\frac{z+x}{xz}\times\frac{z+x}{1}\)(2)
Từ (1) và (2),ta có: vế 1 = vế 1; mà\(\frac{y^2+xz}{y}< y+\frac{xz}{y}< x+z\)
Suy ra điều phải chứng minh...
a) \(\frac{x+1}{2x+6}\)+\(\frac{2x+3}{x\left(x+3\right)}\)
= \(\frac{x+1}{2\left(x+3\right)}\)+ \(\frac{2x+3}{x\left(x+3\right)}\)
= \(\frac{x\left(x+1\right)}{2x\left(x+3\right)}\)+ \(\frac{2\left(2x+3\right)}{2x\left(x+3\right)}\)
= \(\frac{x^2+x+4x+6}{2x\left(x+3\right)}\)
= \(\frac{x^2+5x+6}{2x\left(x+3\right)}\)
= \(\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}\)
= \(\frac{x+2}{2x}\)
b) \(\frac{x-1}{x}\)+ \(\frac{x+2}{2}\)
= \(\frac{2\left(x-1\right)}{2x}\)+ \(\frac{x\left(x+2\right)}{2x}\)
= \(\frac{2x-2+x^2+2x}{2x}\)
= \(\frac{x^2+4x-2}{2x}\)
c) \(\frac{1}{x+y}\)+ \(\frac{-1}{x-y}\)+ \(\frac{2x}{x^2+y^2}\)
= \(\frac{\left(x-y\right)\left(x^2+y^2\right)}{\left(x^2+y^2\right)\left(x-y\right)\left(x+y\right)}\)+\(\frac{-\left(x+y\right)\left(x^2+y^2\right)}{\left(x^2+y^2\right)\left(x-y\right)\left(x+y\right)}\)+ \(\frac{2x\left(x-y\right)\left(x+y\right)}{\left(x^2+y^2\right)\left(x-y\right)\left(x+y\right)}\)
= \(\frac{x^3+xy^2-x^2y-y^3-x^3-xy^2-xy^2-y^3+2x^3+2x^2y-2x^2y+2xy^2}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{2x^3+xy^2-x^2y-2y^3}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{\left(2x^3-2y^3\right)-\left(x^2y-xy^2\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{2\left(x-y\right)\left(x^2+xy+y^2\right)-xy\left(x-y\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{\left(x-y\right)\left(2x^2+2xy+2y^2-xy\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{2x^2+xy+2y^2}{\left(x+y\right)\left(x^2+y^2\right)}\)
e) = \(\frac{3x^2-6xy+3y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
= \(\frac{3\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
=\(\frac{3x-3y}{x^2+xy+y^2}\)
( Mình bận rồi, lát làm câu d nhé)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\ge\frac{\left(1+1+1\right)^2}{x+y+y+z+x+z}\)
\(=\frac{\left(1+1+1\right)^2}{2\left(x+y+z\right)}=\frac{9}{2\left(x+y+z\right)}\)
\(\Rightarrow VT=\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\)
\(\ge\left(x+y+z\right)\cdot\frac{9}{2\left(x+y+z\right)}=\frac{9}{2}=VP\)
Xảy ra khi \(x=y=z\)