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\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)
\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)
Ta có: \(\frac{x^2}{x^4+yz}\le\frac{x^2}{2\sqrt{x^4.yz}}=\frac{x^2}{2x^2\sqrt{yz}}=\frac{1}{2\sqrt{yz}}\)(BĐt cosi) (1)
CMTT: \(\frac{y^2}{y^4+xz}\le\frac{1}{2\sqrt{xz}}\) (2)
\(\frac{z^2}{z^4+xy}\le\frac{1}{2\sqrt{xy}}\)(3)
Từ (1); (2) và (3) =>A = \(\frac{x^2}{x^4+yz}+\frac{y^2}{y^4+xz}+\frac{z^2}{z^4+xy}\le\frac{1}{2}\left(\frac{1}{\sqrt{xz}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xy}}\right)\)
Áp dụng bđt \(ab+bc+ac\le a^2+b^2+c^2\)
cmt đúng: <=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)(luôn đúng)
Khi đó: A \(\le\frac{1}{2}\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{2}\cdot\frac{xy+yz+xz}{xyz}\le\frac{1}{2}\cdot\frac{x^2+y^2+z^2}{xyz}=\frac{3xyz}{2xyz}=\frac{3}{2}\)
http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%AAn-b%C3%ACnh-thu%E1%BA%ADn-2016-2017/
\(VT=\frac{x^2}{x^3-xyz-2013x}+\frac{y^2}{y^3-xyz-2013y}+\frac{z^2}{z^3-xyz-2013z}\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz-2013\left(x+y+z\right)}\)
\(=\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3+3\left[\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\right]}=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}\)=VP
đúng rồi ạ nhưng chỉ cần c/m đẳng thức phụ như thế này thôi ạ\(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\) =>\(\frac{\left(a+b\right)2}{x+y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) hay \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) là xong
Theo giả thiết xy + yz + zx = 1 nên ta có: \(VT=\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+z^2}=\frac{1}{xy+yz+zx+x^2}+\frac{1}{xy+yz+zx+y^2}+\frac{1}{xy+yz+zx+z^2}=\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+x\right)\left(y+z\right)}+\frac{1}{\left(z+x\right)\left(z+y\right)}=\frac{2\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)Theo bất đẳng thức Cauchy-Schwarz: \(\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)^2\le\left(x+y+z\right)\left(\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}\right)=\left(x+y+z\right)\left(\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(y+z\right)\left(y+x\right)}+\frac{z}{\left(z+x\right)\left(z+y\right)}\right)=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)\(\Rightarrow\frac{2}{3}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)^3\le\frac{4\left(x+y+z\right)}{3\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)\)Ta cần chứng minh: \(\frac{2\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\ge\frac{4\left(x+y+z\right)}{3\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)\)
hay \(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\le\frac{3}{2}\)
Bất đẳng thức cuối đúng theo AM - GM do: \(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{y+z}.\frac{y}{x+y}}+\sqrt{\frac{z}{z+x}.\frac{z}{z+y}}\le\frac{\left(\frac{x}{x+y}+\frac{x}{x+z}\right)+\left(\frac{y}{y+z}+\frac{y}{x+y}\right)+\left(\frac{z}{z+x}+\frac{z}{z+y}\right)}{2}=\frac{3}{2}\)Đẳng thức xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
\(\frac{x}{3-yz}+\frac{y}{3-zx}+\frac{z}{3-xy}\le\frac{x}{3-\frac{y^2+z^2}{2}}+\frac{y}{3-\frac{z^2+x^2}{2}}+\frac{z}{3-\frac{x^2+y^2}{2}}\)
\(=\frac{2x}{3+x^2}+\frac{2y}{3+y^2}+\frac{2z}{3+z^2}\le\frac{2x}{4\sqrt[4]{x^2}}+\frac{2y}{4\sqrt[4]{y^2}}+\frac{2z}{4\sqrt[4]{z^2}}\)
\(=\frac{\sqrt{x}}{2}+\frac{\sqrt{y}}{2}+\frac{\sqrt{z}}{2}\le\frac{x^2+3}{8}+\frac{y^2+3}{8}+\frac{z^2+3}{8}\)
\(=\frac{3}{8}+\frac{9}{8}=\frac{3}{2}\)
cách khác: cũng đến chỗ <= sigma 2x/3+x^2
<= 2x/2(x+1) (do x^2+3=x^2+1+2>=2x+2) <= sigma x/x+1 = 3- sigma (1/x+1)
sigma 1/x+1 >= 9/x+y+z+3 dễ rồi
Sử dụng bđt \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)^2\ge3\left(\frac{xy}{z}.\frac{yz}{x}+\frac{yz}{x}.\frac{zx}{y}+\frac{zx}{y}.\frac{xy}{z}\right)=3\left(x^2+y^2+z^2\right)=3\)
\(\Rightarrow\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\ge\sqrt{3}\)