$\frac{1}{1+\sqrt{1+x}}+\frac{3x}{2(1+\sqrt{1+3x})}+\frac{1}{1+\sqrt{1+5x}}=\frac{2\sqrt{1-x^{2}}+\sqrt{1+x}+\sqrt{1-x}}{4}$
Ta có $\sqrt{1-x^2} \le 1$, $\sqrt{1-x}+\sqrt{1+x} \le \sqrt{2(1-x+1+x)}=2$
Từ đó suy ra $VP \le 1$

Lại có $VT \ge \frac{4}{2+\sqrt{x+1}+\sqrt{5x+1}}+\frac{3x}{2(\sqrt{3x+1}+1)}$
Dễ dàng chứng minh $\sqrt{5x+1}+\sqrt{x+1} \le 2\sqrt{3x+1}$
Suy ra $VT \ge \frac{3x+4}{2(1+\sqrt{3x+1})}$
Lại có $3x+4 \ge 2(1+\sqrt{3x+1})$
  (cm bằng cách biến đổi tương đương hoặc dùng bdt AM-GM)
Suy ra $VT \ge 1 \ge VP$
Dấu bẳng xảy ra $\Leftrightarrow x=0$ nên $x=0$ là nghiệm duy nhất của pt

Bạn cần đăng nhập để có thể gửi đáp án

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