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Lời giải:
Áp dụng BĐT AM-GM:
$1=xy+yz+xz+2xyz\leq \frac{(x+y+z)^2}{3}+2.\frac{(x+y+z)^3}{27}$
$\Leftrightarrow 1\leq \frac{t^2}{3}+\frac{2t^3}{27}$ (đặt $x+y+z=t$)
$\Leftrightarrow 2t^3+9t^2-27\geq 0$
$\Leftrightarrow (t+3)^2(2t-3)\geq 0$
$\Leftrightarrow 2t-3\geq 0$
$\Leftrightarrow t\geq \frac{3}{2}$ hay $x+y+z\geq \frac{3}{2}$ (đpcm)
Dấu "=" xảy ra khi $x=y=z=\frac{1}{2}$
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áp dụng bđt cô si ta có:
\(xy\le\frac{x^2+y^2}{2};yz\le\frac{y^2+z^2}{2};zx\le\frac{z^2+x^2}{2}\)
\(\Rightarrow A\ge\sqrt{\frac{x^2+y^2}{2}}+\sqrt{\frac{y^2+z^2}{2}}+\sqrt{\frac{z^2+x^2}{2}}\)
theo bunhia thì \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2;2\left(y^2+z^2\right)\ge\left(y+z\right)^2;2\left(z^2+x^2\right)\ge\left(z+x\right)^2\)
\(\Rightarrow A\ge\sqrt{\frac{\left(x+y\right)^2}{4}}+\sqrt{\frac{\left(y+z\right)^2}{4}}+\sqrt{\frac{\left(z+x\right)^2}{4}}=\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=x+y+z=1\)
Vậy \(Min_A=1\Leftrightarrow x=y=z=\frac{1}{3}\)
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\(VT=6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(=6\left(x+y+z\right)^2-2\left(xy+yz+xz\right)+2\frac{9}{2x+y+z+x+2y+z+x+y+2z}\)
\(\ge6\left(x+y+z\right)^2-2\frac{\left(x+y+z\right)^2}{3}+2\frac{9}{4\left(x+y+z\right)}\)
\(=\: 6\cdot\left(\frac{3}{4}\right)^2-2\cdot\frac{\left(\frac{3}{4}\right)^2}{3}+2\cdot\frac{9}{4\cdot\frac{3}{4}}=9\)
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\(\frac{3}{xy+yz+xz}+\frac{3}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{3}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{3}\right)^2}{x^2+y^2+z^2+2xy+yz+xz}=\frac{\left(\sqrt{6}+\sqrt{3}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{3}\right)^2\)
(*) ta CM :\(\left(\sqrt{6}+\sqrt{3}\right)^2>14\)
TA có \(\left(\sqrt{6}+\sqrt{3}\right)^{^2}=6+3+2\sqrt{18}=9+6\sqrt{2}>9+5=14\)
=> \(\frac{3}{xy+yz+xz}+\frac{3}{x^2+y^2+z^2}>14\)
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Do x nguyên dương
TH1:x=1 Giả sử y=<z
PT<=>2(y+z)=yz-1<=>...<=>(y-2)(z-2)=5
Giải pt nghiệm nguyên dương được nghiệm (1;3;7)
TH2:x>=2
2(y+z)>=2(yz-1)
<=>yz-y-z =<1
<=>(y-1)(z-1) =<2 (1)
Do y,z nguyên dương nên y-1 và z-1 lớn hơn hoặc =0
=>(y-1)(z-1)>=0
Kết hợp với (1) có (y-1)(z-1)=0
hoặc (y-1)(z-1)=1
hoặc (y-1)(z-1)=2
Giải các pt nghiệm nguyên trên ta
KL: pt có các nghiệm (3;5;1),(6;2;1),(4;3;1),(3;1;5),(6;1;2),
(4;1;3),(2;2;3),(2;3;2),(1;3;7),(1;7;3...
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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 \(x=y=z=\frac{1}{\sqrt{3}}\)