ta dễ chứng minh được \(x+y\ge\frac{2\sqrt{2}}{5}-\frac{2}{5}\)\(\Rightarrow\)\(x+y+\frac{2\sqrt{2}}{5}-\frac{2}{5}>0\)
\(P=\frac{5\left(x+y+\frac{2\sqrt{2}}{5}-\frac{2}{5}\right)\left(\frac{5}{2}\left(x+y-\left(\frac{2\sqrt{2}}{5}-\frac{2}{5}\right)\right)\left(\frac{5}{2}\left(x+y\right)+\sqrt{2}+1\right)-\frac{9}{4}\left(x-y\right)^2\right)}{\frac{5}{2}\left(x+y\right)+\sqrt{2}+1}\)
\(+\left(\frac{\frac{45}{2}\left(x+y+\frac{2\sqrt{2}}{5}-\frac{2}{5}\right)}{5\left(x+y\right)+\sqrt{2}+1}+\frac{9}{2}\right)\left(x-y\right)^2+6-4\sqrt{2}\ge6-4\sqrt{2}\)
Dấu "=" xảy ra khi \(x=y=\frac{\sqrt{2}-1}{5}\)
Ta chứng minh: \(P\ge6-4\sqrt{2}+\left(2-\sqrt{2}\right)\left(4x^2+4y^2+17xy+5x+5y-11\right)\)
Hay là:
\(\frac{\left(9+4\sqrt{2}\right)\left(98x-298y-130+225\sqrt{2}y+85\sqrt{2}\right)^2}{9604}+\frac{18\left(2\sqrt{2}-1\right)\left(-5y-1+\sqrt{2}\right)^2}{36+16\sqrt{2}}\ge0\)
Việc còn lại là của mọi người.
Dòng đầu là \(P\ge6-4\sqrt{2}+\left(2-\sqrt{2}\right)\left(4x^2+4y^2+17xy+5x+5y-1\right)\)!
Em đánh dư./
Ta có a=x+y; \(xy\le\frac{a^2}{4};1\le4x^2-4y^2-17xy-5x-5y\le5a+4a^2+\frac{9}{4}a^2\)
\(\Rightarrow\left(1+\frac{5}{2}a\right)^2\ge2\Leftrightarrow a\ge\frac{2}{5}\left(\sqrt{2}-1\right)\)Lúc đó ta có:
\(P=17x^2+17y^2+16xy\ge17a^2-4,5a^2=\frac{25}{2}a^2\ge2\left(\sqrt{2}-1\right)=6-4\sqrt{2}\)
Dấu "=" xảy ra <=> \(x=y=\frac{1}{5}\left(\sqrt{2}-1\right)\)
