# Prove that: $2^a+3^b<3a+4b$

Let be $a, b$ in $(0,1)$ such that $a+b>1$. I need to prove that:

$$2^a+3^b<3a+4b$$

I'm looking for an elementary proof that doesn't resort to the calculus tools.

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Are you allow to use the fact that $2^a + 3^b$ is convex? Then you just need to check at the end-points $(a,b) \in \{ (0,1), (1,0), (1,1)\}$. –  Willie Wong Jun 15 '12 at 12:36
@Willie Wong: yes. This idea came to me too late. –  Chris's sis Jun 15 '12 at 12:41

From the graph of the function $f(x)=2^x$ we see that on interval $(0,1)$ it is bellow the line $y=x+1$ joining the points $(0,f(0))=(0,1)$ and $(1,f(1))=(1,2)$. Thus we have $$2^a<1+a$$ for $a\in(0,1)$.

Using similar argument for $3^x$ we get $$3^b<1+2b$$ for $b\in(0,1)$.

Adding the two inequalities together and using $1<a+b$ we obtain $$2^a+3^b<2+a+2b<2(a+b)+a+2b=3a+4b.$$

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nice proof! Thanks! –  Chris's sis Jun 15 '12 at 12:41
Using Bernoulli's Inequality, $2^a \leq 1 + a$ and $3^b \leq 1 + 2b$. Therefore, $$2^a + 3^b \leq 2 + a + 2b < 2(a + b) + a + 2b = 3a + 4b$$
The problem is that both proofs use calculus in a hidden way ;) Proving Bernoulli for $a \in (0,1)$ requires calculus, and so does the concavity of $2^x$.... –  N. S. Jun 15 '12 at 14:18
That is true, but I think convexity of $2^x$ can be established without calculus. $2^{\lambda x + (1 - \lambda)y} = (2^x)^{\lambda} (2^y)^{1 - \lambda} \leq \lambda 2^x + (1 - \lambda) 2^y$ where the last step comes from the generalized AM-GM inequality. –  TenaliRaman Jun 15 '12 at 16:54