1
$\begingroup$

let $ a,b \in (0,1)$ ,$a+b>1$

then : prove that : $$2^a+3^b<3a+4b$$

my try :

$$f(x):=e^x\\ e^{(2^a+4^b)}

now ?

  • 1
    The hypothesis are contradictory... please fix. – 2017-02-22
  • 0
    Is my edit what you meant? – 2017-02-22
  • 0
    @S.C.B. Thank . – 2017-02-22
  • 1
    You've got a strictly convex function on a triangular domain, so you can just check that it's true (non strictly) at the three vertices. – 2017-02-22

2 Answers 2

3

Let $T$ be the triangular region $\{(a,b)\in\mathbb{R}^2: a,b\in[0,1], a+b\geq 1\}$.
$T$ is closed and convex. The function $$ f(a,b) = (2^a-3a)+(3^b-4b) $$ is convex as the sum of two convex functions, hence $\max_{(a,b)\in T}f(a,b)$ is attained at one of the vertices of $T$. We have $f(1,0)=f(0,1)=0$ and $f(1,1)=-2$, hence $f$ is less than $0$ on the interior of $T$, as wanted.

0

By Bernoulli $$2^a+3^b=(1+1)^a+(1+2)^b\leq1+a\cdot1+1+b\cdot2=$$ $$=2+a+2b<2(a+b)+a+2b=3a+4b$$

  • 0
    That is true, but the same solution was already given in an answer to the "duplicate" question. – 2017-02-22
  • 0
    @Martin R 13 Now I see it. Thank you! – 2017-02-22