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\(\frac{4x^2y^2}{\left(x^2+y^2\right)^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge3\)
\(\Leftrightarrow\frac{4x^2y^2}{\left(x^2+y^2\right)^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}-3\ge0\)
\(\Leftrightarrow\frac{4x^4y^4+x^4\left(x^2+y^2\right)^2+y^4\left(x^2+y^2\right)^2-3x^2y^2\left(x^2+y^2\right)^2}{x^2y^2\left(x^2+y^2\right)^2}\)
\(\Leftrightarrow4x^4y^4+x^4\left(x^4+2x^2y^2+y^4\right)+y^4\left(x^4+2x^2y^2+y^4\right)-3x^2y^2\left(x^4+2x^2y^2+y^4\right)\ge0\)
\(\Leftrightarrow4x^4y^4+x^8+2x^6y^2+x^4y^4+2x^2y^6+y^8-3x^6y^2-6x^4y^4-3x^2y^6\ge0\)
\(\Leftrightarrow x^8+y^8-x^6y^2-x^2y^6\ge0\)
\(\Leftrightarrow x^6\left(x^2-y^2\right)-y^6\left(x^2-y^2\right)\ge0\)
\(\Leftrightarrow\left(x^2-y^2\right)^2\left(x^4+x^2y^2+y^4\right)\ge0\) ( luôn đúng )
\(\Rightarrow\frac{4x^2y^2}{\left(x^2+y^2\right)^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow x=y\)
\(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\)
\(\Leftrightarrow\frac{x^4+y^4+4x^2y^2}{x^2y^2}\ge\frac{3x^3y+3y^3x}{x^2y^2}\)
\(\Leftrightarrow x^4+y^4+4x^2y^2-3x^3y-3xy^3\ge0\)
\(\Leftrightarrow x^2\left(x^2-2xy+y^2\right)+y^2\left(x^2-2xy+y^2\right)-x^3y-xy^3+2x^2y^2\ge0\)
\(\Leftrightarrow\left(x^2+y^2\right)\left(x^2-2xy+y^2\right)-xy\left(x^2+y^2-2xy\right)\ge0\Leftrightarrow\left(x^2-xy+y^2\right)\left(x-y\right)^2\ge0\)(đúng)
\(\Rightarrowđpcm."="\Leftrightarrow x=y\)
Tham khảo ở đây nha bạn!
http://olm.vn/hoi-dap/question/520851.html
Đặt \(\dfrac{x}{y}+\dfrac{y}{x}=a\)\(\Rightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2=a^2\Rightarrow\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}=a^2-2\)
Ta có \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4=a^2-2+4=a^2+2\)
\(3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)=3a\)
Ta có \(a^2+2-3a=a^2-2.a.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{1}{4}=\left(a-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\)
lạ có \(\dfrac{x}{y}+\dfrac{y}{x}-2=\dfrac{x^2}{xy}-\dfrac{2xy}{xy}+\dfrac{y^2}{xy}=\dfrac{\left(x-y\right)^2}{xy}\ge0\)
\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}\ge2\)\(\Rightarrow a\ge2\Rightarrow a-\dfrac{3}{2}\ge\dfrac{1}{2}\)\(\Rightarrow\left(a-\dfrac{3}{2}\right)^2\ge\dfrac{1}{4}\Rightarrow\left(a-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge0\)
\(\Rightarrow a^2+2-3a\ge0\Rightarrow a^2+2\ge3a\Rightarrow\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
\(\left\{{}\begin{matrix}x;y>0\\\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\end{matrix}\right.\) \(\begin{matrix}\left(1\right)\\\left(2\right)\end{matrix}\)
từ (2) có \(\Leftrightarrow\left(\dfrac{x^2}{y^2}+2.\dfrac{x}{y}.\dfrac{y}{x}+\dfrac{y^2}{x^2}\right)+2-3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge0\)
\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2-3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+2\ge0\)
\(\Leftrightarrow\left[\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\right]-\left[\left(\dfrac{x}{y}+\dfrac{y}{x}\right)-2\right]\ge0\)
\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}-2\right)\left(\dfrac{x}{y}+\dfrac{y}{x}-1\right)\ge0\) (3)
từ (1) có \(\dfrac{x}{y}+\dfrac{y}{x}=\left(\sqrt{\dfrac{x}{y}}-\sqrt{\dfrac{y}{x}}\right)^2+2\ge2\) (4)
từ (4) ; \(\left\{{}\begin{matrix}\left(\dfrac{x}{y}+\dfrac{y}{x}-1\right)>0\\\dfrac{x}{y}+\dfrac{y}{x}-2\ge0\end{matrix}\right.\) (I)
từ (I) => (3) đúng mọi phép biến đổi là <=> đẳng thức khi \(\dfrac{x}{y}=\dfrac{y}{x}\Rightarrow x=y\)=> dpcm
\(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2+2\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2-3\left(\frac{y}{x}+\frac{x}{y}\right)+2\ge0\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-1\right)\left(\frac{x}{y}+\frac{y}{x}-2\right)\ge0\)
\(\Leftrightarrow\frac{\left(x^2-xy+y^2\right)\left(x^2-2xy+y^2\right)}{x^2y^2}\ge0\)
\(\Leftrightarrow\frac{\left[\left(x-\frac{1}{2}y\right)^2+\frac{3}{4}y^2\right]\left(x-y\right)^2}{x^2y^2}\ge0\) ( đúng )
Vậy đẳng thức đã được chứng minh .
Dấu \("="\) xảy ra khi \(x=y\)
DƯƠNG PHAN KHÁNH DƯƠNG: Dùng AM-GM cũng được mà
Áp dụng BĐT AM-GM ta có:\(\left\{{}\begin{matrix}\frac{x^2}{y^2}+1\ge2.\frac{x}{y}\\\frac{y^2}{x^2}+1\ge2.\frac{y}{x}\\\frac{x}{y}+\frac{y}{x}\ge2\end{matrix}\right.\)
Dấu " = " xảy ra <=> x=y
\(\Rightarrow\frac{x^2}{y^2}+1+\frac{y^2}{x^2}+1+2\ge2\left(\frac{x}{y}+\frac{y}{x}\right)+2\)
Có: \(2\left(\frac{x}{y}+\frac{y}{x}\right)+2-3\left(\frac{x}{y}+\frac{y}{x}\right)=\left(\frac{x}{y}+\frac{y}{x}\right)\left(2-3\right)+2\ge2.\left(-1\right)+2=0\)\(\Rightarrow\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\)
Dấu " = " xảy ra <=> x=y
1/ Với số dương ta luôn có \(\frac{x}{y}+\frac{y}{x}\ge2\) (Cauchy hoặc quy đồng chuyển vế sẽ chứng minh được dễ dàng). Ta cần chứng minh:
\(\frac{x^2}{y^2}+\frac{y^2}{x^2}+2.\frac{x}{y}.\frac{y}{x}+2\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2+2\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\) (1)
Đặt \(\frac{x}{y}+\frac{y}{x}=a\ge2\) thì (1) trở thành:
\(a^2+2\ge3a\Leftrightarrow a^2-3a+2\ge0\Leftrightarrow\left(a-1\right)\left(a-2\right)\ge0\) (2)
Do \(a\ge2\Rightarrow\left\{{}\begin{matrix}a-1>0\\a-2\ge0\end{matrix}\right.\Rightarrow\left(a-1\right)\left(a-2\right)\ge0\)
\(\Rightarrow\left(2\right)\) đúng, vậy BĐT được chứng minh. Dấu "=" xảy ra khi \(x=y\)
2/ \(B=\left(x^2-2x\right)\left(y^2+6y\right)+12\left(x^2-2x\right)+3\left(y^2+6y\right)+2045\)
\(B=\left(x^2-2x\right)\left(y^2+6y+12\right)+3\left(y^2-6y+12\right)-36+2045\)
\(B=\left(x^2-2x+3\right)\left(y^2+6y+12\right)+2009\)
\(B=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]+2009\)
Do \(\left\{{}\begin{matrix}\left(x-1\right)^2+2\ge2\\\left(y+3\right)^2+3\ge3\end{matrix}\right.\)
\(\Rightarrow B\ge2.3+2009=2015\)
\(\Rightarrow B_{min}=2015\) khi \(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
\(BĐT\Leftrightarrow\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}+2\right)-3\left(\frac{x}{y}+\frac{y}{x}\right)+2\ge0\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2-3\left(\frac{x}{y}+\frac{y}{x}\right)+2\ge0\)
\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-1\right)\left(\frac{x}{y}+\frac{y}{x}-2\right)\ge0\) (Luôn đúng vì \(\frac{x}{y}+\frac{y}{x}\ge2\forall x;y>0\))