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Lời giải:
Ta có:
\(\text{VT}=\frac{1}{x^2+y^2+2}+\frac{1}{y^2+z^2+2}+\frac{1}{z^2+x^2+2}\)
\(\Rightarrow 2\text{VT}=\frac{2}{x^2+y^2+2}+\frac{2}{y^2+z^2+2}+\frac{2}{z^2+x^2+2}\)
\(2\text{VT}=1-\frac{x^2+y^2}{x^2+y^2+2}+1-\frac{y^2+z^2}{y^2+z^2+2}+1-\frac{z^2+x^2}{z^2+x^2+2}\)
\(2\text{VT}=3-\left(\frac{x^2+y^2}{x^2+y^2+2}+\frac{y^2+z^2}{y^2+z^2+2}+\frac{z^2+x^2}{z^2+x^2+2}\right)=3-A\)
Áp dụng BĐT Cauchy-Schwarz:
\(A\geq \frac{(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2}{2(x^2+y^2+z^2)+6}=\frac{(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2}{2(x^2+y^2+z^2+xy+yz+xz)}(*)\)
Xét tử số:
\((\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2\)
\(=2(x^2+y^2+z^2)+2(\sqrt{(x^2+y^2)(x^2+z^2)}+\sqrt{(x^2+y^2)(y^2+z^2)}+\sqrt{(y^2+z^2)(z^2+x^2)})\)
Áp dụng BĐT Bunhiacopxky:
\(\sqrt{(x^2+y^2)(x^2+z^2)}\geq \sqrt{(x^2+yz)^2}=x^2+yz\)
\(\sqrt{(x^2+y^2)(y^2+z^2)}\geq \sqrt{(xz+y^2)^2}=xz+y^2\)
\(\sqrt{(y^2+z^2)(z^2+x^2)}\geq \sqrt{(z^2+xy)^2}=z^2+xy\)
\(\Rightarrow \sum \sqrt{(x^2+y^2)(x^2+z^2)}\geq x^2+y^2+z^2+xy+yz+xz\)
\(\Rightarrow (\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2\geq 4(x^2+y^2+z^2)+2(xy+yz+xz)\)
\(\geq 3(x^2+y^2+z^2)+3(xy+yz+xz)=3(x^2+y^2+z^2+xy+yz+xz)\)
(theo BĐT AM-GM)
Do đó: Từ \((*)\Rightarrow A\geq \frac{3(x^2+y^2+z^2+xy+yz+xz)}{2(x^2+y^2+z^2+xy+yz+xz)}=\frac{3}{2}\)
\(\Rightarrow 2\text{VT}\leq 3-\frac{3}{2}=\frac{3}{2}\)
\(\Rightarrow \text{VT}\leq \frac{3}{4}\) (đpcm)
Dấu bằng xảy ra khi \(x=y=z=1\)
We have: \(\dfrac{1}{x^2+y^2+2}=\dfrac{1}{x^2+y^2+z^2+2-z^2}\le\dfrac{1}{5-z^2}\)
Similarly and by adding them:
\(\dfrac{1}{5-x^2}+\dfrac{1}{5-y^2}+\dfrac{1}{5-z^2}\le\dfrac{3}{4}\left(\circledast\right)\)
We know that \(\dfrac{1}{5-x^2}\le\dfrac{3\left(x^2+x\right)}{8\left(x^2+x+1\right)}\)
\(\Leftrightarrow-\dfrac{\left(x-1\right)^2\left(3x^2+9x+8\right)}{8\left(x^2-5\right)\left(x^2+x+1\right)}\le0\) It's obviously
\(\Rightarrow L.H.S_{\left(\circledast\right)}\le\dfrac{3}{8}\left(\dfrac{x^2+x}{x^2+x+1}+\dfrac{y^2+y}{y^2+y+1}+\dfrac{z^2+z}{z^2+z+1}\right)\le\dfrac{3}{4}\)
The equality occur when \(x=y=z=1\)
Done!
\(VT=\sum\frac{x}{\sqrt{1+x^2}}=\sum\frac{x}{\sqrt{xy+xz+yz+x^2}}=\sum\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\frac{1}{2}\sum\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)\(\Rightarrow VT\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+z}+\frac{y}{x+y}+\frac{z}{x+z}+\frac{z}{y+z}\right)=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
\(\dfrac{1}{{xy}} + \dfrac{2}{{{x^2} + {y^2}}} = \dfrac{2}{{2xy}} + \dfrac{2}{{{x^2} + {y^2}}} = 2\left( {\dfrac{1}{{2xy}} + \dfrac{1}{{{x^2} + {y^2}}}} \right) \ge 2\dfrac{4}{{2{x^2} + 2xy + {y^2}}} = \dfrac{8}{{{{\left( {x + y} \right)}^2}}} = 8\)
Đẳng thức xảy ra khi \(x=y=\dfrac{1}{2}\)
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Bài 1:
Áp dụng BĐT AM-GM:
\(9=x+y+xy+1=(x+1)(y+1)\leq \left(\frac{x+y+2}{2}\right)^2\)
\(\Rightarrow 4\leq x+y\)
Tiếp tục áp dụng BĐT AM-GM:
\(x^3+4x\geq 4x^2; y^3+4y\geq 4y^2\)
\(\frac{x}{4}+\frac{1}{x}\geq 1; \frac{y}{4}+\frac{1}{y}\geq 1\)
\(\Rightarrow x^3+y^3+x^2+y^2+5(x+y)+\frac{1}{x}+\frac{1}{y}\geq 5(x^2+y^2)+\frac{3}{4}(x+y)+2\)
Mà:
\(5(x^2+y^2)\geq 5.\frac{(x+y)^2}{2}\geq 5.\frac{4^2}{2}=40\)
\(\frac{3}{4}(x+y)\geq \frac{3}{4}.4=3\)
\(\Rightarrow A= x^3+y^3+x^2+y^2+5(x+y)+\frac{1}{x}+\frac{1}{y}\geq 40+3+2=45\)
Vậy \(A_{\min}=45\Leftrightarrow x=y=2\)
Bài 2:
\(B=\frac{a^2}{a-1}+\frac{2b^2}{b-1}+\frac{3c^2}{c-1}\)
\(B-24=\frac{a^2}{a-1}-4+\frac{2b^2}{b-1}-8+\frac{3c^2}{c-1}-12\)
\(=\frac{a^2-4a+4}{a-1}+\frac{2(b^2-4b+4)}{b-1}+\frac{3(c^2-4c+4)}{c-1}\)
\(=\frac{(a-2)^2}{a-1}+\frac{2(b-2)^2}{b-1}+\frac{3(c-2)^2}{c-1}\geq 0, \forall a,b,c>1\)
\(\Rightarrow B\geq 24\)
Vậy \(B_{\min}=24\Leftrightarrow a=b=c=2\)
\(P=\frac{\sqrt{386}^2}{x^2+y^2+z^2}+\frac{\sqrt{700}^2}{2\left(xy+yz+zx\right)}\ge\frac{\left(\sqrt{386}+\sqrt{700}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{386}+\sqrt{700}\right)^2\)
Bây giờ chỉ cần chứng minh:
\(\left(\sqrt{386}+\sqrt{700}\right)^2>2015\)
Ta có \(\left(\sqrt{386}+\sqrt{700}\right)^2>\left(\sqrt{361}+\sqrt{676}\right)^2=2025>2015\) (đpcm)