Proving inequality for all real $a$ and $b$. I am trying to prove the following two inequalities
My first intuition was to take the natural log of both sides, then simplify them to get an obvious statement regarding a and b. But I can't reduce the natural log of the right hand-side of the inequality (because of the addition operator).
I'm feeling really stumped and dumb, so a hint would be great!
 A: for the first one:
dividing by $$e^{a+2b}\ne 0$$ and we get
$$3\le e^{2(a-b)}+\frac{2}{e^{a-b}}$$
substituting $$z=e^{a-b}$$ then we get
$$3\le z^2+\frac{2}{z}$$
can you finish?
and this is equivalent to $$0\le (z+2)(z-2)^2$$ which is true. For your send question:
we get the general case : $$\frac{a^n+b^n}{2}\geq \left(\frac{a+b}{2}\right)^n$$
A: The first inequality:
By AM-GM
$$\frac{e^{3a}+2e^{3b}}{3}\geq\sqrt[3]{e^{3a}\left(e^{3b}\right)^2}=e^{a+2b}$$
The second inequality it's Jensen for $f(x)=x^{2018}$:
$$\frac{a^{2018}+b^{2018}}{2}\geq\left(\frac{a+b}{2}\right)^{2018}.$$
A: If we divide first one with $e^{a+2b}$ we get $$3\leq (e^{a-b})^2+2e^{b-a}$$
Now let $x=e^{a-b}$ so we have to prove $$3\leq x^2+{2\over x}$$ 
which is true by inequality between arithmetic and geometric mean:
$$ x^2+{1\over x}+{1\over x} \geq 3\cdot \sqrt[3]{x^2{1\over x}{1\over x}}=3$$
A: *

*Use AM-GM on $e^{3a},e^{3b},e^{3b}$ so that $$\frac{e^{3a}+2e^{3b}}{3} \ge e^{a}e^{2b}$$

*Use Jensen's inequality on $x\mapsto x^{2018}$, which is a convex function (since it has a positive second derivative except at $x=0$).
\begin{align}
\left(\frac{a+b}{2}\right)^{2018} &\le \frac12 a^{2018}+\frac12 b^{2018} \\
(a+b)^{2018} &\le 2^{2017} a^{2018} + 2^{2017} b^{2018}
\end{align}



Since OP starts by taking log, but none of the existing answers use log, I'll continue OP's work.


*

*Take log on both sides to get


\begin{align}
e^{a+2b} &\le \frac13 e^{3a} + \frac23 e^{3b} \\
\iff a+2b &\le \log \left( \frac13 e^{3a} + \frac23 e^{3b} \right) \\
\iff \frac13 \log e^{3a} + \frac23 \log e^{3b} &\le \log \left( \frac13 e^{3a} + \frac23 e^{3b} \right) \\
\iff \frac13 f(e^{3a}) + \frac23 f(e^{3b}) &\le f \left( \frac13 e^{3a} + \frac23 e^{3b} \right),
\end{align}
which is Jensen's inequality for the concave function $f: x \mapsto \log x$.
