cái này mình nhằm. thay vì bằng 1 nó bằng 11/13 nha. giúp mình với cảm ơn nhiều

áp dụng bdt amgm ta có  \(xyz\le\left(\frac{x+y+z}{3}\right)^3=\frac{1}{3^3}=\frac{1}{27}\)

 \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\le\left(\frac{x+y+y+z+x+z}{3}\right)^3=\left(\frac{2\left(x+y+z\right)}{3}\right)^3=\frac{8}{27}\)

\(\Rightarrow xyz\left(x+y\right)\left(y+z\right)\left(x+z\right)\le\frac{1}{27}.\frac{8}{27}=\left(\frac{2}{9}\right)^3\)

dau = xay ra khi x=y=z=1/3 

ta có \(x^4+y^4\ge2x^2y^2\)               \(y^4+z^4\ge2y^2z^2\) \(z^4+x^4\ge2x^2z^2\)

\(\Rightarrow2\left(x^4+y^4+z^4\right)\ge2\left(x^2y^2+y^2z^2+z^2x^2\right)\)\(\Rightarrow x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\)

mat khac \(\left(a^2+b^2+c^2\right)\ge\frac{\left(a+b+c\right)^2}{3}\) (tu cm)

\(\Rightarrow x^2y^2+y^2z^2+z^2x^2\ge\frac{\left(xy+yz+zx\right)^2}{3}=\frac{1}{3}\)

min =1/3 \(\) dau = xay ra khi \(x=y=z=\frac{+-\sqrt{3}}{3}\)

xyz=1

=>\(\left|x\right|\cdot\left|y\right|\cdot\left|z\right|=1\)

Theo AM-GM, ta được:

\(\vert{}xy\vert{} + \vert{}yz\vert{} + \vert{}zx\vert{} \ge 3 \sqrt[3]{\vert{}xy\vert{} \cdot \vert{}yz\vert{} \cdot \vert{}zx\vert{}} = 3 \sqrt[3]{(xyz)^2} = 3 \cdot 1 = 3\)

Dấu '=' xảy ra khi |x|=|y|=|z|=1

Đặt \(Q=15\sqrt{x^2 + y^2 + z^2}-7(x+y-z)\)

Theo Cauchy, ta có:

\(1 \cdot x + 1 \cdot y + 1 \cdot (-z) \le \sqrt{(1^2 + 1^2 + 1^2)(x^2 + y^2 + (-z)^2)} = \sqrt{3(x^2 + y^2 + z^2)}\)

=>\(x + y - z \le \sqrt{3}\sqrt{x^2 + y^2 + z^2}\)

=>\(-7(x+y-z)\ge-7\sqrt{3}\sqrt{x^2 + y^2 + z^2}\)

Do đó, ta có:

\(Q \ge 15\sqrt{x^2 + y^2 + z^2} - 7\sqrt{3}\sqrt{x^2 + y^2 + z^2} = (15 - 7\sqrt{3})\sqrt{x^2 + y^2 + z^2}\)

=>\(Q \ge (15 - 7\sqrt{3})\sqrt{3} = 15\sqrt{3} - 21\)

Dấu '=' xảy ra khi \(\begin{cases}x=y=-z\\ \vert{}x\vert{}=\vert{}y\vert{}=\vert{}z\vert{}=1\\ xyz=1\end{cases}\iff x=1,y=1,z=-1\)

\(P=(\vert{}xy\vert{}+\vert{}yz\vert{}+\vert{}zx\vert{})Q+1\)

=>\(P\ge3\cdot(15\sqrt{3}-21)+1=45\sqrt{3}-63+1=45\sqrt{3}-62\)

Dấu '=' xảy ra khi x=y=z=1

\(\left|xy\right|+\left|yz\right|+\left|zx\right|\)

\(A=\frac{1}{\sqrt{x^2-xy+y^2}}+\frac{1}{\sqrt{y^2-yz+z^2}}+\frac{1}{\sqrt{z^2-zx+x^2}}\)

\(=\frac{1}{\sqrt{\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y-z\right)^2+\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z-x\right)^2+\frac{1}{2}\left(z^2+x^2\right)}}\)

\(\le\frac{1}{\sqrt{\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z^2+x^2\right)}}\)

\(\le\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)