Proving inequality for all real $a$ and $b$.

Question

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!

Answer 1

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 : $$\frac{a^n+b^n}{2}\geq \left(\frac{a+b}{2}\right)^n$$

Answer 2

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 $f(x)=x^{2018}$: $$\frac{a^{2018}+b^{2018}}{2}\geq\left(\frac{a+b}{2}\right)^{2018}.$$

Answer 3

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:

$$ x^2+{1\over x}+{1\over x} \geq 3\cdot \sqrt[3]{x^2{1\over x}{1\over x}}=3$$

Answer 4

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

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

1. 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 $f: x \mapsto \log x$.