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Lời giải :
\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2\ge a^2x^2+2abxy+b^2y^2\)
\(\Leftrightarrow a^2y^2-2abxy+b^2x^2\ge0\)
\(\Leftrightarrow\left(ay-bx\right)^2\ge0\)( luôn đúng )
Dấu "=" xảy ra \(\Leftrightarrow\frac{a}{x}=\frac{b}{y}\)
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\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(=a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)
\(\left(ax+by+cz\right)^2\)
\(=c^2z^2+2bcyz+2acxz+b^2y^2+2abxy+a^2x^2\)
\(\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\)\(\ge\left(ax+by+cz\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)
\(\ge c^2z^2+2bcyz+2acxz+b^2y^2+2abxy+a^2x^2\)
\(\Leftrightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2\)
\(\ge2bcyz+2acxz+2abxy\)
\(\Leftrightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2\)\(-2bcyz-2acxz-2abxy\ge0\)
\(\Leftrightarrow\left(a^2y^2-2abxy+b^2x^2\right)+\left(a^2z^2-2acxz+c^2x^2\right)\)
\(+\left(b^2z^2-2bcyz+c^2y^2\right)\ge0\)
\(\Leftrightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2\ge0\)
(Điều trên đúng vì \(\hept{\begin{cases}\left(ay-bx\right)^2\ge0\\\left(az-cx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\end{cases}}\))
Vậy\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\) \(\ge\left(ax+by+cz\right)^2\)
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Dat \(A=\frac{x^4+y^4}{x^4-y^4}-\frac{xy}{x^2-y^2}+\frac{x+y}{2\left(x-y\right)}\)
\(=\frac{2x^4+2y^4-2xy\left(x^2+y^2\right)+\left(x+y\right)^2\left(x^2+y^2\right)}{2x^4-2y^4}\)
\(=\frac{2x^4+2y^4+\left(x^2+y^2\right)\left[\left(x+y\right)^2-2xy\right]}{2x^4-2y^4}\)
\(=\frac{2x^4+2y^4+\left(x^2+y^2\right)^2}{2x^4-2y^4}\)
\(\Rightarrow A\ge\frac{2x^4+x^4}{2x^4}=\frac{3}{2}\)
\(\Rightarrow P=2017A\ge2017.\frac{3}{2}=\frac{6051}{2}\)
Dau '=' xay ra khi \(y=0\)
Ta có:\(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
\(\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow\left(x-y\right)^2\ge0\)(đúng) \(\Rightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)(1)
Mặt khác: \(\frac{\left(x+y\right)^2}{2}\ge2xy\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow x^2+y^2\ge2xy\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\)(đúng) \(\Rightarrow\frac{\left(x+y\right)^2}{2}\ge2xy\) (2)
Từ (1) và (2) => đpcm