Chứng minh rằng: \(\frac{a^2+1}{4b^2}+\frac{b^2+1}{4c^2}+\frac{c^2+1}{4a^2}\geq \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\)

Ta có: \(VT=\left ( \frac{a^{2}}{4b^{2}}+\frac{1}{4b^{2}} \right )+\left ( \frac{b^{2}}{4c^{2}}+\frac{1}{4c^{2}} \right )+\left ( \frac{c^{2}}{4a^{2}}+\frac{1}{4a^{2}} \right )\)

\(\geq \frac{a}{2b^{2}}+\frac{b}{2c^{2}}+\frac{c}{2a^{2}}=\frac{1}{2}\left ( \frac{a}{b^{2}}+\frac{b}{c^{2}}+\frac{c}{a^{2}} \right )\)

Mặt khác: \(\frac{a}{b^{2}}+\frac{1}{a}\geq \frac{2}{b};\; \; \; \; \frac{b}{c^{2}}+\frac{1}{b}\geq \frac{2}{c};\; \; \; \; \frac{c}{a^{2}}+\frac{1}{c}\geq \frac{2}{a}\)

Cộng theo vế các BĐT trên ta được: \(\frac{a}{b^{2}}+\frac{b}{c^{2}}+\frac{c}{a^{2}}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Suy ra:

\(VT\geq \frac{1}{2}\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )=\frac{1}{4}\left [ \left ( \frac{1}{a}+\frac{1}{b} \right )+\left ( \frac{1}{b}+\frac{1}{c} \right )+\left ( \frac{1}{c}+\frac{1}{a} \right ) \right ]\)

\(\geq \frac{1}{4}\left [ \frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a} \right ]=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=VP\)

Đẳng thức xảy ra khi và chỉ khi: a = b = c = 1