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Cái đầu tiên là \(\sqrt[n]{\frac{a_1^n+a_2^n+a_3^n+...+a_n^n}{n}}\)nhé.
Lời giải:
Áp dụng BĐT AM-GM ta có:
\(\text{VT}=x-\frac{x}{x^2+z}+y-\frac{y}{y^2+x}+z-\frac{z}{z^2+y}=(x+y+z)-\left(\frac{x}{x^2+z}+\frac{y}{y^2+x}+\frac{z}{z^2+y}\right)\)
\(\geq (x+y+z)-\left(\frac{x}{2\sqrt{x^2z}}+\frac{y}{2\sqrt{y^2x}}+\frac{z}{2\sqrt{z^2y}}\right)=(x+y+z)-\frac{1}{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)(1)\)
Từ giả thiết \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
Cauchy-Schwarz:
\(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3(2)\)
\(\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2\leq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(1+1+1)=9\)
\(\Rightarrow \left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)\leq 3(3)\)
Từ \((1);(2);(3)\Rightarrow \text{VT}\geq 3-\frac{1}{2}.3=\frac{3}{2}\)
Mặt khác: \(\text{VP}=\frac{1}{2}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{2}\)
Do đó \(\text{VT}\geq \text{VP}\) (đpcm)
Dấu "=" xảy ra khi $x=y=z=1$
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 \(x=y=z=\frac{1}{\sqrt{3}}\)
Theo AM-GM: \(x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\)
\(\Rightarrow\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\)
Tương tự: \(\frac{2\sqrt{y}}{y^3+z^2}\le\frac{1}{yz}\)
\(\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{zx}\)
Cộng vế với vế => \(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)
Theo AM-GM; \(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}}{2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Dấu " = " xảy ra <=> x=y=z=1
Áp dụng bất đẳng thức Cacuhy - Schwarz
\(\Rightarrow\hept{\begin{cases}x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\\y^3+z^2\ge2\sqrt{y^3z^2}=2yz\sqrt{y}\\z^3+x^2\ge2\sqrt{z^3x^2}=2xz\sqrt{z}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\\\frac{2\sqrt{y}}{y^3+z^2}\le\frac{2\sqrt{y}}{2yz\sqrt{y}}=\frac{1}{yz}\\\frac{2\sqrt{z}}{z^3+x^2}\le\frac{2\sqrt{z}}{2xz\sqrt{z}}=\frac{1}{xz}\end{cases}}\)
\(\Rightarrow VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(1\right)\)
Áp dụng bất đẳng thức Cacuchy Schwarz
\(\Rightarrow\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2y^2}}=\frac{2}{xy}\\\frac{1}{y^2}+\frac{1}{z^2}\ge2\sqrt{\frac{1}{y^2z^2}}=\frac{2}{yz}\\\frac{1}{z^2}+\frac{1}{x^2}\ge2\sqrt{\frac{1}{z^2x^2}}=\frac{2}{xz}\end{cases}}\)
\(\Rightarrow2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\ge2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow VT\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(\Leftrightarrow\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\left(đpcm\right)\)
Ta có:
y02 + ay0 + b = 0
\(\Leftrightarrow\)y04 = (ay0 + b)2
\(\le\)(a2 + b2)(y02 + 1)
\(\Rightarrow\)y04 - 1 < (a2 + b2)(y02 + 1)
\(\Rightarrow\)y02 - 1 < a2 + b2
\(\Rightarrow\)y02 < 1 + a2 + b2
3/ Dễ thấy \(0\le x,y,z\le1\)
Ta có:
x2 + y2 + z2 = x3 + y3 + z3
\(\Leftrightarrow\)x2(1 - x) + y2(1 - y) + z2(1 - z) = 0
Dấu = xảy ra khi (x, y, z) = (0,0,1) và các hoán vị của nó