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Đặt \(P=\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Do x,y,z là các số thực dương nên ta biến đổi \(P=\frac{1}{\sqrt{1+\frac{1}{x^2}}}+\frac{1}{\sqrt{1+\frac{1}{y^2}}}+\frac{1}{\sqrt{1+\frac{1}{z^2}}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Đặt \(a=\frac{1}{x^2};b=\frac{1}{y^2};c=\frac{1}{z^2}\left(a,b,c>0\right)\)thì \(xy+yz+zx=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}=1\)và \(P=\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}}+a+b+c\)
Biến đổi biểu thức P=\(\left(\frac{1}{2\sqrt{a+1}}+\frac{1}{2\sqrt{a+1}}+\frac{a+1}{16}\right)+\left(\frac{1}{2\sqrt{b+1}}+\frac{1}{2\sqrt{b+1}}+\frac{b+1}{16}\right)\)\(+\left(\frac{1}{2\sqrt{c+1}}+\frac{1}{2\sqrt{c+1}}+\frac{c+1}{16}\right)+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{b}-\frac{3}{16}\)
Áp dụng Bất Đẳng Thức Cauchy ta có
\(P\ge3\sqrt[3]{\frac{a+1}{64\left(a+1\right)}}+3\sqrt[3]{\frac{b+1}{64\left(b+1\right)}}+3\sqrt[3]{\frac{c+1}{64\left(c+1\right)}}+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{16}-\frac{3}{16}\)
\(=\frac{33}{16}+\frac{15}{16}\left(a+b+c\right)\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{abc}\)
Mặt khác ta có \(1=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\ge3\sqrt[3]{\frac{1}{abc}}\Leftrightarrow abc\ge27\)
\(\Rightarrow P\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{27}=\frac{33}{16}+\frac{15}{16}\cdot9=\frac{21}{2}\)
Dấu "=" xảy ra khi a=b=c hay \(x=y=z=\frac{\sqrt{3}}{3}\)
\(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)=9\Rightarrow x+y+z\ge3\)
\(P=\sum\frac{x^2}{\sqrt{x^3+8}}=\sum\frac{x^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}}\ge\sum\frac{2x^2}{x^2-x+6}\ge\frac{2\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)
\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{x^2+y^2+z^2+6-\left(x+y+z\right)+12}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)+12}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)+12}-1+1\)
\(\Rightarrow P\ge\frac{\left(x+y+z\right)^2+\left(x+y+z\right)-12}{\left(x+y+z\right)^2-\left(x+y+z\right)+12}+1=\frac{\left(x+y+z-3\right)\left(x+y+z+4\right)}{\left(x+y+z\right)^2-\left(x+y+z\right)+12}+1\)
Do \(x+y+z-3\ge0\Rightarrow P\ge1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
Gọi \(A=\sum\dfrac{x^3}{\sqrt{y^2+3}}\)
Theo Holder: \(A.A.\left(\left(y^2+3\right)+\left(z^2+3\right)+\left(x^2+3\right)\right)\ge\left(x^3+y^3+z^3\right)^3\)
\(\Rightarrow A^2\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+9}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}=\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+xy+yz+zx}\ge\dfrac{\left(x^3+y^3+z^3\right)^3}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}\)
Ta có đánh giá sau: \(x^3+y^3+z^3\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x+y+z}\ge\dfrac{\left(x+y+z\right)^3}{9}\)
\(\Rightarrow A^2\ge\dfrac{\dfrac{\left(x+y+z\right)^3}{9}}{\left(x+y+z\right)^2+\dfrac{\left(x+y+z\right)^2}{3}}=\dfrac{x+y+z}{12}\ge\dfrac{\sqrt{3\left(xy+yz+zx\right)}}{12}\ge\dfrac{1}{4}\)
\(\Rightarrow A\ge\dfrac{1}{2}\)
Tương tự, ta được:
\(\left(2-y\right)\left(2-z\right)>=\dfrac{\left(x+1\right)^2}{4}\)
và \(\left(2-z\right)\left(2-x\right)>=\left(\dfrac{y+1}{2}\right)^2\)
=>8(2-x)(2-y)(2-z)>=(x+1)(y+1)(z+1)
(x+yz)(y+zx)<=(x+y+yz+xz)^2/4=(x+y)^2*(z+1)^2/4<=(x^2+y^2)(z+1)^2/4
Tương tự, ta cũng co:
\(\left(y+xz\right)\left(z+y\right)< =\dfrac{\left(y^2+z^2\right)\left(x+1\right)^2}{2}\)
và \(\left(z+xy\right)\left(x+yz\right)< =\dfrac{\left(z^2+x^2\right)\left(y+1\right)^2}{2}\)
Do đó, ta được:
\(\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)< =\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
=>ĐPCM
\(\sqrt{\frac{xy}{xy+z}}=\sqrt{\frac{xy}{xy+z\left(x+y+z\right)}}=\sqrt{\frac{xy}{\left(x+z\right)\left(y+z\right)}}\le\frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)\)
Tương tự: \(\sqrt{\frac{yz}{yz+x}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{z}{x+z}\right)\) ; \(\sqrt{\frac{zx}{zx+y}}\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{z}{y+z}\right)\)
Cộng vế với vế ta có đpcm
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
min hay max bạn
Mk nghĩ là x3,y3,z3.
Áp dụng BĐT AM-GM:
\(\Sigma_{cyc}\left(\frac{x^2}{\sqrt{x^3+8}}\right)=\Sigma_{cyc}\left(\frac{x^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}}\right)\)\(\ge2\Sigma_{cyc}\left(\frac{x^2}{x^2-x+6}\right)\)
Áp dụng BĐT Cauchy-Schwart:
\(2\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)\(=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-2\left(xy+yz+zx\right)-\left(x+y+z\right)+18}\)\(\ge\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-2\left(x+y+z\right)-\left(x+y+z\right)+18}\)
gt\(\Leftrightarrow3\left(x+y+z\right)\le3\left(xy+yz+zx\right)\le\left(x+y+z\right)^2\)
\(\Leftrightarrow\left(x+y+z\right)^2-3\left(x+y+z\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x+y+z\le0\\x+y+z\ge3\end{matrix}\right.\)
Đặt t=x+y+z\(\left(t\ge3\right)\)
Cần c/m:\(\frac{2t^2}{t^2-3t+18}\ge1\)
Có :\(t^2-3t+18>0\)
\(\Rightarrow2t^2\ge t^2-3t+18\)
\(\Leftrightarrow t^2+3t-18\ge3^2+3.3-18=0\)(Đúng)
Vậy min =1
Dấu = xra khi x=y=z=1.
#Walker
Kiểm tra giùm em đúng ko ạ Akai Haruma