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\(\Sigma\frac{x^3}{y^2}=\Sigma\frac{x}{y^2}\left(x-y\right)^2+\frac{\Sigma z\left(x^3-yz^2\right)^2}{xyz\left(x+y+z\right)}+\Sigma\frac{x^2}{y}\ge\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\)
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 \(x=y=z=\frac{1}{\sqrt{3}}\)
Ta có : \(\frac{2}{x^2+y^2}+\frac{2}{y^2+z^2}+\frac{2}{z^2+x^2}=\frac{x^2+y^2+z^2}{x^2+y^2}+\frac{x^2+y^2+z^2}{y^2+z^2}+\frac{x^2+y^2+z^2}{z^2+x^2}=\frac{z^2}{x^2+y^2}+\frac{x^2}{y^2+z^2}+\frac{y^2}{z^2+x^2}+3\)
Ta lại có : \(x^2+y^2\le2xy\Leftrightarrow\frac{z^2}{x^2+y^2}\le\frac{z^2}{2xy}\)
\(y^2+z^2\le2yz\Leftrightarrow\frac{x^2}{y^2+z^2}\le\frac{x^2}{2yz}\)
\(z^2+x^2\le2zx\Leftrightarrow\frac{y^2}{z^2+x^2}\le\frac{y^2}{2zx}\)
Cộng vế theo vế ta có :
\(\frac{z^2}{x^2+y^2}+\frac{x^2}{y^2+z^2}+\frac{y^2}{z^2+x^2}\le\frac{z^2}{2xy}+\frac{x^2}{2yz}+\frac{y^2}{2zx}\)
\(\Leftrightarrow\frac{z^2}{x^2+y^2}+\frac{x^2}{y^2+z^2}+\frac{y^2}{z^2+x^2}+3\le\frac{z^2}{2xy}+\frac{x^2}{2yz}+\frac{y^2}{2zx}+3\)
\(\Leftrightarrow\frac{2}{x^2+y^2}+\frac{2}{y^2+z^2}+\frac{2}{z^2+x^2}\le\frac{x^2+y^2+z^2}{2xyz}+3\)
\(\Rightarrowđpcm\)
Áp dụng BĐT Cô-si dạng Engel,ta có :
\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}\)
Mà \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le x+y+z\)
\(\Rightarrow\)\(\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\ge\frac{3}{2}\)
Dấu "=" xảy ra khi x = y = z = \(\frac{3}{2}\)
Ta có \(\frac{x^3}{\left(y+z\right)^2}=\frac{x^3}{\left(2018-x\right)^2}\)
Xét \(\frac{x^3}{\left(2018-x\right)^2}\ge x-\frac{1009}{2}\)
<=> \(x^3\ge\left(2018^2-2.2018.x+x^2\right)\left(x-\frac{1009}{2}\right)\)
<=> \(x^3\ge x^3-x^2\left(\frac{1009}{2}+2018.2\right)+x\left(2018.1009+2018^2\right)-\frac{2018^2.1009}{2}\)
<=> \(\frac{9081}{2}x^2-6.1009^2.x+2018.1009^2\ge0\)
<=> \(\frac{9081}{2}\left(x^2-\frac{2.2018}{3}.x+\left(\frac{2018}{3}\right)^2\right)\ge0\)
<=> \(\frac{9081}{2}\left(x-\frac{2018}{3}\right)^2\ge0\)( luôn đúng)
=> \(\frac{x^3}{\left(y+z\right)^2}\ge x-\frac{1009}{2}\)
Khi đó \(VT\ge x-\frac{1009}{2}+y-\frac{1009}{2}+z-\frac{1009}{2}=2018-\frac{3}{2}.1009=\frac{1009}{2}\)(ĐPCM)
Dấu bằng xảy ra khi \(x=y=z=\frac{2018}{3}\)
Ta có : \(\frac{x^3}{\left(y+z\right)^2}=\frac{x^3}{\left(2018-x\right)^2}\)
xét \(\frac{x^3}{\left(2018-x\right)^2}\ge x-\frac{1009}{2}\)
<=> \(x^3\ge\left(x^2-2.2018.x+2018^2\right)\left(x-\frac{1009}{2}\right)\)
<=> \(x^3\ge x^3-x^2\left(\frac{1009}{2}+2.2018\right)+x\left(2018^2+1009.2018\right)-\frac{2018^2.1009}{2}\ge0\)
<=> \(\frac{9081}{2}x^2-6.1009^2.x+2018.1009^2\ge0\)
<=> \(\frac{9081}{2}.\left(x-\frac{2018}{3}\right)^2\ge0\)( luôn đúng)
=> \(\frac{x^3}{\left(y+z\right)^2}\ge x-\frac{1009}{2}\)
Khi đó \(P\ge x+y+z-\frac{3.1009}{2}=\frac{1009}{2}\)(ĐPCM)
Dấu bằng xảy ra khi \(x=y=z=\frac{2018}{3}\)
Bằng một số bước tính toán cơ bản, chúng ta có được:
\(VT-VP=\Sigma_{cyc}\frac{x\left(x-z\right)^2}{2\left(x^2+z^2\right)}\ge0\)
tth_old : t chán cái kiểu SOS gì đó của m rồi đấy.