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[hoanganhtuan96vn]

Chứng minh rằng: $$\sqrt{1+\sqrt{1-x^{2}}}\left [ \sqrt{1+x^{3}}-\sqrt{1-x^{3}} \right ]\leq 2\sqrt{2}+\sqrt{2+2x^{2}}$$

Edited by Nguyen Lam Thinh, 07-02-2013 - 23:01.

[phanquockhanh]

$\sqrt{1+\sqrt{1-x^{2}}}\left [ \sqrt{1+x^{3}}-\sqrt{1-x^{3}} \right ]\leq 2\sqrt{2}+\sqrt{2+2x^{2}}$

Đề bài có chút vấn đề.Hình như VP là $2\sqrt{2}+\sqrt{2-2x^{2}}$ mới phải

Đk:$\left | x \right |\leq 1$.Ta đặt:x=cost,$t\in$[0;$\pi$].Khi đó:
(ycbt)$\Leftrightarrow \sqrt{1+sint}\left ( \sqrt{(1+cost)^{3}} -\sqrt{(1-cost)^{3}}\right )\leq \sqrt{2}(2+sint)$
$\Leftrightarrow \sqrt{(sin\frac{t}{2}+cos\frac{t}{2})^{2}}\left [ \sqrt{(2cos^{2}\frac{t}{2})^{3}}-\sqrt{(2sin^{2}\frac{t}{2})^{3}}\right ]\leq \sqrt{2}(2+sint)$
$\Leftrightarrow 2\sqrt{2}(cos^{2}\frac{t}{2}-sin^{2}\frac{t}{2})(1+sin\frac{t}{2}cos\frac{t}{2})\leq \sqrt{2}(2+sint)$
$\Leftrightarrow cost.(2+sint)\leq 2+sint$
$\Leftrightarrow cost\leq 1$ (đúng)

Edited by phanquockhanh, 08-02-2013 - 18:21.