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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.

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    @Willie Wong: yes. This idea came to me too late.2012-06-15

2 Answers 2

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

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

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    @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 :P2012-06-15