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Với mọi x;y;z ta luôn có:

\(\left(x+y-1\right)^2+\left(z-\dfrac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow x^2+y^2+2xy-2x-2y+1+z^2-z+\dfrac{1}{4}\ge0\)

\(\Leftrightarrow x^2+y^2+z^2+\dfrac{5}{4}+2xy-2x-2y-z\ge0\)

\(\Leftrightarrow2+2xy-2x-2y\ge z\)

\(\Leftrightarrow2\left(1-x\right)\left(1-y\right)\ge z\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=\dfrac{1}{2}\)

\(\frac{x^2}{y^2}+\frac{y^2}{z^2}\ge2\cdot\sqrt{\frac{x^2}{y^2}\cdot\frac{y^2}{z^2}}=2\cdot\sqrt{\frac{x^2}{z^2}}=2\cdot\frac{x}{z}\)

\(\frac{y^2}{z^2}+\frac{z^2}{x^2}\ge2\cdot\sqrt{\frac{y^2}{z^2}\cdot\frac{z^2}{x^2}}=2\cdot\sqrt{\frac{y^2}{x^2}}=2\cdot\frac{y}{x}\)

\(\frac{z^2}{x^2}+\frac{x^2}{y^2}\ge2\cdot\sqrt{\frac{z^2}{x^2}\cdot\frac{x^2}{y^2}}=2\cdot\sqrt{\frac{z^2}{y^2}}=2\cdot\frac{z}{y}\)

Do đó; \(\left(\frac{x^2}{y^2}+\frac{y^2}{z^2}\right)+\left(\frac{y^2}{z^2}+\frac{z^2}{x^2}\right)+\left(\frac{z^2}{x^2}+\frac{x^2}{y^2}\right)\ge2\cdot\frac{x}{z}+2\cdot\frac{y}{x}+2\cdot\frac{z}{y}\)

=>\(2\left(\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}\right)\ge2\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)\)

=>\(\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}\ge\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\)

\(\frac{x^2}{y^2}+\frac{y^2}{z^2}\ge2\cdot\sqrt{\frac{x^2}{y^2}\cdot\frac{y^2}{z^2}}=2\cdot\sqrt{\frac{x^2}{z^2}}=2\cdot\frac{x}{z}\)

\(\frac{y^2}{z^2}+\frac{z^2}{x^2}\ge2\cdot\sqrt{\frac{y^2}{z^2}\cdot\frac{z^2}{x^2}}=2\cdot\sqrt{\frac{y^2}{x^2}}=2\cdot\frac{y}{x}\)

\(\frac{z^2}{x^2}+\frac{x^2}{y^2}\ge2\cdot\sqrt{\frac{z^2}{x^2}\cdot\frac{x^2}{y^2}}=2\cdot\sqrt{\frac{z^2}{y^2}}=2\cdot\frac{z}{y}\)

Do đó; \(\left(\frac{x^2}{y^2}+\frac{y^2}{z^2}\right)+\left(\frac{y^2}{z^2}+\frac{z^2}{x^2}\right)+\left(\frac{z^2}{x^2}+\frac{x^2}{y^2}\right)\ge2\cdot\frac{x}{z}+2\cdot\frac{y}{x}+2\cdot\frac{z}{y}\)

=>\(2\left(\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}\right)\ge2\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)\)

=>\(\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}\ge\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\)

x + y + z = 0 ⇒ x 3 + y 3 + z 3 = 3 x y z ⇒ ( x 3 + y 3 + z 3 ) ( x 2 + y 2 + z 2 ) = 3 x y z ( x 2 + y 2 + z 2 ) ⇒ x 5 + y 5 + z 5 + x 2 y 2 ( x + y ) + y 2 z 2 ( y + z ) + z 2 x 2 ( z + x ) = 3 x y z ( x 2 + y 2 + z 2 ) ⇒ x 5 + y 5 + z 5 − x y z ( x y + y x + z x ) = 3 x y z ( x 2 + y 2 + z 2 ) ⇒ 2 ( x 5 + y 5 + z 5 ) = 5 x y z ( x 2 + y 2 + z 2)

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