1.By Cauchy-Schwarz we have:

\(\left(1+1\right)\left(a^4+b^4\right)\ge\left(a^2+b^2\right)^2\)

Need prove \(\dfrac{\left(a^2+b^2\right)^2}{2}\ge a^2+b^2\Leftrightarrow\left(a^2+b^2\right)^2\ge2\left(a^2+b^2\right)\)

\(\Leftrightarrow a^2+b^2\ge2\). By C-S again :

\(\left(1+1\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\Leftrightarrow a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}=2\)

It's right or we are done :3

2. Remark : \(1+a^2=ab+bc+ca+a^2=\left(a+b\right)\left(a+c\right)\)

\(1+b^2=ab+bc+ca+b^2=\left(b+a\right)\left(b+c\right)\)

\(1+c^2=ab+bc+ca+c^2=\left(c+a\right)\left(c+b\right)\)

Using Cauchy for these positive numbers \(\dfrac{1}{a+b},\dfrac{1}{b+c},\dfrac{1}{c+a}\)

We have : \(\dfrac{1}{a+b}+\dfrac{1}{b+c}\ge\dfrac{2}{\sqrt{\left(a+b\right)\left(b+c\right)}}\Rightarrow\dfrac{b}{a+b}+\dfrac{b}{b+c}\ge\dfrac{2b}{\sqrt{1+b^2}}\)

\(\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{2}{\sqrt{\left(b+c\right)\left(c+a\right)}}\Rightarrow\dfrac{c}{b+c}+\dfrac{c}{c+a}\ge\dfrac{2c}{\sqrt{1+c^2}}\)

\(\dfrac{1}{c+a}+\dfrac{1}{a+b}\ge\dfrac{2}{\sqrt{\left(c+a\right)\left(a+b\right)}}\Rightarrow\dfrac{a}{c+a}+\dfrac{a}{a+b}\ge\dfrac{2a}{\sqrt{1+a^2}}\)

Plus each pair of inequalities we have :

\(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\ge2\left(\dfrac{a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}\right)\)

\(\Rightarrow\dfrac{3}{2}\ge\dfrac{a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}\)

2. By Cauchy-Schwarz we have:

\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(b+c\right)}}\)

\(\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)

Similarly :\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{b+c}\right)\)

\(L.H.S\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{3}{2}=R.H.S\)

Yes, thanks Egst for your answers.

I also have another ideas for these math problems :

1. Using Bunhiacopxki, we have :

\(2\left(a^2+b^2\right)\ge\left(a+b^2\right)\)

\(\Rightarrow2\left(a^2+b^2\right)\ge4\Rightarrow a^2+b^2\ge2\)

We have : \(\left(a^2-1\right)^2\ge0\Rightarrow a^4+1\ge2a^2\)

\(\left(b^2-1\right)^2\ge0\Rightarrow b^4+1\ge2b^2\)

Plus each pair of the two inequalities we have :

\(a^4+b^4+2\ge2\left(a^2+b^2\right)\)

\(\Rightarrow a^4+b^4\ge a^2+b^2+\left(a^2+b^2-2\right)\ge a^2+b^2\) because \(a^2+b^2\ge2\)

So \(a^4+b^4\ge a^2+b^2\)

Anyone who has different answer plz comment in this question ;D