Question

[Momozono Nanami]

Cho a,b,c,d là những số thực không âm t/m

\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2\)

CMR

\(\sqrt{\frac{a^2+1}{2}}+\sqrt{\frac{b^2+1}{2}}+\sqrt{\frac{c^2+1}{2}}+\sqrt{\frac{d^2+1}{2}}\ge3\left(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)-8\)

Answer 1

[Kiệt Nguyễn]

Ta sẽ chứng minh: \(\sqrt{\frac{x^4+1}{2}}+\frac{4x^2}{x^2+1}\ge3x\)

Thật vậy: \(\Leftrightarrow\left(\sqrt{\frac{x^4+1}{2}}-x\right)+2\left(\frac{2x^2}{x^2+1}-x\right)\ge0\)

\(\Leftrightarrow\left(x-1\right)^2\left[\frac{\left(x+1\right)^2}{2\sqrt{\frac{x^4+1}{2}}+2x}-\frac{2x}{x^2+1}\right]\ge0\)

Bây giờ ta quy về chứng minh: \(\frac{\left(x+1\right)^2}{2\sqrt{\frac{x^4+1}{2}}}\ge\frac{2x}{x^2+1}\Leftrightarrow\left(x^2+1\right)\left(x+1\right)^2\ge4x\left(\sqrt{\frac{x^4+1}{2}+x}\right)\)

\(\Leftrightarrow x^4+1+2x^3+2x\ge2x^2+4x\sqrt{\frac{x^4+1}{2}}\)

\(\Leftrightarrow\frac{x^4+1}{2}+x^3+x\ge x^2+2x\sqrt{\frac{x^4+1}{2}}\)

Bất đẳng thức trên đúng theo AM - GM:

\(\frac{x^4+1}{2}+x^3+x\ge\left(\frac{x^4+1}{2}+x^2\right)+x^2\ge2x\sqrt{\frac{x^4+1}{2}}+x^2\)

Vậy hoàn tất chứng minh trên nên ta có:

\(\sqrt{\frac{a^2+1}{2}}+\frac{4a}{a+1}\ge3\sqrt{a}\);\(\sqrt{\frac{b^2+1}{2}}+\frac{4b}{b+1}\ge3\sqrt{b}\)

\(\sqrt{\frac{c^2+1}{2}}+\frac{4c}{c+1}\ge3\sqrt{c}\); \(\sqrt{\frac{d^2+1}{2}}+\frac{4c}{d+1}\ge3\sqrt{d}\)

Cộng từng vế của các bđt trên. ta được: \(\text{Σ}_{cyc}\sqrt{\frac{a^2+1}{2}}\ge3\left(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)\)

\(-4\left(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\right)\)\(=3\left(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)-8\)

Dấu "=" xảy ra khi a = b = c = 1