Chọn C
Với các số thực \(m,\, \, n dương, x,\, \, y \)tùy ý , ta có:
\(\left(x+y\right)^{2} =\left(\sqrt{m} .\frac{x}{\sqrt{m} } +\sqrt{n} .\frac{y}{\sqrt{n} } \right)^{2} \le \left(m+n\right)\left(\frac{x^{2} }{m} +\frac{y^{2} }{n} \right)\) (Bất đẳng thức Bunhiacopski)
Hay: \(\frac{x^{2} }{m} +\frac{y^{2} }{n} \ge \frac{\left(x+y\right)^{2} }{m+n} .\)
Do đó: \(\frac{a^{2} }{-b} +\frac{4b^{2} }{a} \ge \frac{\left(a-2b\right)^{2} }{a-b} . \)
Suy ra \(\left(\frac{a^{2} }{b} -\frac{4b^{2} }{a} \right)\left(b-a\right)^{2} \le \left(a-2b\right)^{2} \left(b-a\right).\)
Mặt khác :\(\left(7+5\sqrt{2} \right)\left(ab-a^{2} \right)\left[4\left(\sqrt{2} +1\right)b+a\right]=\left(1+\sqrt{2} \right)^{3} a\left(b-a\right)\left[4\left(\sqrt{2} +1\right)b+a\right]\)
\(=\left(1+\sqrt{2} \right)\left(a-b\right)\left[\left(3+2\sqrt{2} \right)a\right]\left[-4\left(\sqrt{2} +1\right)b-a\right]\)
\(\le \left(1+\sqrt{2} \right)\left(a-b\right)\left(\frac{\left(3+2\sqrt{2} \right)a-4\left(\sqrt{2} +1\right)b-a}{2} \right)^{2}\) (Bất đẳng thức Cô si).
\(=\left(1+\sqrt{2} \right)\left(a-b\right)\left(1+\sqrt{2} \right)^{2} \left(a-2b\right)^{2} =\left(7+5\sqrt{2} \right)\left(a-b\right)\left(a-2b\right)^{2} . \)
Suy ra \(8\sqrt{\left(7+5\sqrt{2} \right)\left(ab-a^{2} \right)\left[4\left(\sqrt{2} +1\right)b+a\right]} \le 8\sqrt{\left(7+5\sqrt{2} \right)\left(a-b\right)\left(a-2b\right)^{2} } . \)
Đặt \(t=\sqrt{\left(a-b\right)\left(a-2b\right)^{2} } \left(t>0\right) .\)
Khi đó:\( P\le -t^{2} +8t\sqrt{7+5\sqrt{2} \; } =f\left(t\right) trong đó {\mathop{\max }\limits_{t\in \left(0;+\infty \right)}} f\left(t\right)=16\left(7+5\sqrt{2} \right) \)
hay \(P\le 16\left(7+5\sqrt{2} \right)\)
Do đó: \(\left(5\sqrt{2} -7\right)P\le 16\left(5\sqrt{2} -7\right)\left(5\sqrt{2} +7\right)=16.\)
