Prove: $2^{a}+ 2^{b}+ 2^{c}\leqq 3$ with $\sqrt{2}\left ( a+ b+ c \right )= \sqrt{a^{2}+ 4}+ \sqrt{b^{2}+ 4}+ \sqrt{c^{2}+ 4}$
By AM_GM, we have:
$$\sum_{cyc}\log_2a=\log_2abc\leq\log_2\left(\frac{a+b+c}{3}\right)^3$$
I can't find and I can't continue! Please help! Thanks!
Your second problem is wrong. Try $b=c=0$ and $a=4\sqrt2+2\sqrt5.$
For your first problem, which was $\sum\limits_{cyc}\log_2a\leq3.$ Because $$\sum_{cyc}(1-\log_2a)=\sum_{cyc}\left(1-\log_2a+\frac{1}{\sqrt2\ln2}\left(\sqrt2a-\sqrt{a^2+4}\right)\right)\geq0.$$ Let $$f(a)=1-\log_2a+\frac{1}{\sqrt2\ln2}\left(\sqrt2a-\sqrt{a^2+4}\right).$$ Thus, $$f'(a)=\frac{(a-1)\sqrt{2(a^2+4)}-a^2}{2\ln2a\sqrt{a^2+4}}.$$
We see that $f'(a)<0$ for $0<a<1$, but for $a\geq1$ we obtain:
$$f'(x)=\frac{2(a-1)^2(a^2+4)-a^4}{2a\ln2\sqrt{a^2+4}\left((a-1)\sqrt{2(a^2+4)}+a^2\right)}=$$ $$=\frac{(a-2)(a^3-2a^2+6a-4)}{2a\ln2\sqrt{a^2+4}\left((a-1)\sqrt{2(a^2+4)}+a^2\right)}>0$$ for $a>2$, which gives $f(a)\geq f(2)=0$ and we are done!