Có BĐT sau:

\(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\Leftrightarrow2x^2+2y^2\ge x^2+2xy+y^2\Leftrightarrow\left(x-y\right)^2\ge0\left(true!\right)\)

Ta có:

\(x^8+y^8\ge\frac{\left(x^4+y^4\right)^2}{2}\ge\frac{\left[\frac{\left(x^2+y^2\right)^2}{2}\right]^2}{2}\ge2\)

Dấu "=" xảy ra tại a=b=1

\(A=\frac{y}{x}.\sqrt{\frac{x^2}{\left(y^2\right)^2}}=\frac{y}{x}.\frac{x}{y^2}=\frac{1}{y}< 0.\)

Đơn giản hơn vì:

\(\sqrt{\frac{x^2}{y^4}}>0\); \(\frac{y}{x}< 0\)=> \(A< 0.\)

Ta có: \(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}=\frac{a^2+ab+1}{\sqrt{a^2+ab+2ab+c^2}}\ge\frac{a^2+ab+1}{\sqrt{a^2+ab+a^2+b^2+c^2}}=\sqrt{a^2+ab+1}\)

\(\sqrt{a^2+ab+1}=\sqrt{a^2+ab+a^2+b^2+c^2}=\sqrt{\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2+a^2+c^2}\)

\(=\frac{1}{\sqrt{5}}.\sqrt{\left(\frac{9}{4}+\frac{3}{4}+1+1\right)\left(\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2+a^2+c^2\right)}\)

\(\ge\frac{1}{\sqrt{5}}\sqrt{\left(\frac{3}{2}\left(a+\frac{b}{2}\right)+\frac{3}{2}b+a+c\right)^2}\)

\(=\frac{1}{\sqrt{5}}\left(\frac{5}{2}a+\frac{3}{2}b+c\right)\)

=> \(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}\ge\frac{1}{\sqrt{5}}\left(\frac{5}{2}a+\frac{3}{2}b+c\right)\)

Tương tự ta cũng chứng minh đc:

 \(\frac{b^2+bc+1}{\sqrt{b^2+3bc+a^2}}\ge\frac{1}{\sqrt{5}}\left(\frac{5}{2}b+\frac{3}{2}c+a\right)\)

\(\frac{c^2+ca+1}{\sqrt{c^2+3ca+b^2}}\ge\frac{1}{\sqrt{5}}\left(\frac{5}{2}c+\frac{3}{2}a+b\right)\)

=> \(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}+\frac{b^2+bc+1}{\sqrt{b^2+3bc+a^2}}+\frac{c^2+ca+1}{\sqrt{c^3+3ca+b^2}}\ge\frac{1}{\sqrt{5}}\left(5a+5b+5c\right)\)

\(=\sqrt{5}\left(a+b+c\right)\)

Dấu "=" xảy ra <=> a = b = c =\(\frac{1}{\sqrt{3}}\)

Đặt : \(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)

=> \(A^2=16-2\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)

\(=16-2\sqrt{8^2-4\left(10+2\sqrt{5}\right)}\)

\(=16-2\sqrt{24-8\sqrt{5}}\)

\(=16-2\sqrt{20-2.2\sqrt{5}.2+4}\)

\(=16-2\sqrt{\left(2\sqrt{5}-2\right)^2}\)

\(=16-2\left(2\sqrt{5}-2\right)=20-4\sqrt{5}\)

=> \(A=\sqrt{20-4\sqrt{5}}\)

Có: \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\ge4\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\)

<=> \(\left(\frac{a}{b}+1\right)+\left(\frac{b}{c}+1\right)+\left(\frac{c}{a}+1\right)\ge\frac{4a}{a+b}+\frac{4b}{b+c}+\frac{4c}{c+a}\)

<=> \(\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a}\ge\frac{4a}{a+b}+\frac{4b}{b+c}+\frac{4c}{c+a}\)

<=> \(\frac{a+b}{b}+\frac{b+c}{c}+\frac{c+a}{a}+\frac{4b}{a+b}+\frac{4c}{b+c}+\frac{4a}{c+a}\)

\(\ge\frac{4a}{a+b}+\frac{4b}{b+c}+\frac{4c}{c+a}+\frac{4b}{a+b}+\frac{4c}{b+c}+\frac{4a}{c+a}\)

<=> \(\left(\frac{a+b}{b}+\frac{4b}{a+b}\right)+\left(\frac{b+c}{c}+\frac{4c}{b+c}\right)+\left(\frac{c+a}{a}+\frac{4a}{c+a}\right)\)

\(\ge\left(\frac{4a}{a+b}+\frac{4b}{a+b}\right)+\left(\frac{4b}{b+c}+\frac{4c}{b+c}\right)+\left(\frac{4c}{c+a}+\frac{4a}{c+a}\right)\)

<=> \(\left(\frac{a+b}{b}+\frac{4b}{a+b}\right)+\left(\frac{b+c}{c}+\frac{4c}{b+c}\right)+\left(\frac{c+a}{a}+\frac{4a}{c+a}\right)\ge4+4+4\)

<=> \(\left(\frac{a+b}{b}+\frac{4b}{a+b}\right)+\left(\frac{b+c}{c}+\frac{4c}{b+c}\right)+\left(\frac{c+a}{a}+\frac{4a}{c+a}\right)\ge12\)(1)

Áp dụng Cô-si: (1) đúng.

Vậy Bất đẳng thức ban đầu đúng.

"=" <=> a = b = c.

\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\ge4\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}\right)\)

\(\Leftrightarrow\left(\frac{a}{b}+1\right)+\left(\frac{b}{c}+1\right)+\left(\frac{c}{a}+1\right)\ge4\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\)

\(\Leftrightarrow\frac{a+b}{b}-\frac{4a}{a+b}+\frac{b+c}{c}-\frac{4b}{b+c}+\frac{c+a}{a}-\frac{4c}{c+a}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2}{b\left(a+b\right)}+\frac{\left(b-c\right)^2}{c\left(b+c\right)}+\frac{\left(c-a\right)^2}{a\left(a+c\right)}\ge0\)

Luôn đúng vì a,b,c là các số dương

Dấu "=" xảy ra <=> a=b=c