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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.

user 1591719
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2 Answers2

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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 $$2^a+3^b<2+a+2b<2(a+b)+a+2b=3a+4b.$$

Martin Sleziak
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Using Bernoulli's Inequality, $2^a \leq 1 + a$ and $3^b \leq 1 + 2b$. Therefore, $$2^a + 3^b \leq 2 + a + 2b < 2(a + b) + a + 2b = 3a + 4b$$

TenaliRaman
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  • Oh dear, just realized its the same as the above answer. embarassed – TenaliRaman Jun 15 '12 at 14:03
  • 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
  • True, but as far as I know, the generalized AM-GM inequality can only be proven for rational powers without Calculus ;) Anyhow, your point is probably as much non-calculus as possible for this problem... – N. S. Jun 15 '12 at 19:55
  • @N.S. I was expecting some catch-22 like that :-). But then, we can use the fact that for f(x) continuous, it is enough to show midpoint convexity (thus using AM-GM in a rational power setting). Of course, to show f(x) is continuous we may have to resort to calculus again :P – TenaliRaman Jun 15 '12 at 19:58