$ (1+\sin{x})^{\cos{x}} + (1+\cos{x})^{\sin{x}} > 3x $

Question

How do I show that, for $ 0 < x < \dfrac{\pi}{4} $ (first quadrant), the inequality $ (1+\sin{x})^{\cos{x}} + (1+\cos{x})^{\sin{x}} > 3x $ is valid?

I've tried Bernoulli's, but it took me to a false inequality (though all restrictions were respected). I actually thought of using Jensen's, but I don't know where to begin.

Bernoulli's inequality would give an upper bound for the LHS, not a lower bound, as the exponent is in $(0,1)$. — Macavity, Oct 28 '14 at 03:00

Nathaniel Bubis · Accepted Answer · 2014-10-28T08:18:30.967

5

Use:

$$\cos x > 1/\sqrt{2},\quad \sin x > x - x^3/6>0$$ So that we have: $$(1+\sin{x})^{\cos{x}} + (1+\cos{x})^{\sin{x}} > 1+\left(1+\frac{1}{\sqrt{2}}\right)^{x-x^3/6}$$

Now, we can use the series expansion of $(1+a)^x$: $$(1+a)^x = 1+ \log(1+a)x+\frac{1}{2}\log^2(1+a) x^2 + \ldots = \sum_{n=0}^\infty \frac{1}{n!}\log^n(1+a) x^n$$ Since all members of the series are positive for $a>0$, we can write: $$1+\left(1+\frac{1}{\sqrt{2}}\right)^{x-x^3/6} > 2+\log(1+1/\sqrt2)x>\frac{x}{2}+2>3x$$ With the last being true since $\pi/4<4/5$, the point where $3x=2+x/2$.

edited Oct 28 '14 at 08:18

answered Oct 27 '14 at 21:14

Nathaniel Bubis

33,018

I could not understand your reasonings. Can you please work with more basic statements/definitions?
Sorry if I seem rude, but I don't like "taking rabbits of out hats" for math proofs.
– Dela Corte Oct 27 '14 at 22:39
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@DelaCorte - is that better? – Nathaniel Bubis Oct 28 '14 at 07:57
Way better. I solved it by another (simpler!) way though, but your solution was very helpful. Thanks! – Dela Corte Oct 29 '14 at 15:50
@DelaCorte - welcome to accept it if it solved your problem. – Nathaniel Bubis Oct 29 '14 at 15:51
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Sometimes "proofs by Way 1" help me develop "proofs by Way 2", and even "proofs by my own way". Thanks @nbubis. – Dela Corte Oct 29 '14 at 16:02

$ (1+\sin{x})^{\cos{x}} + (1+\cos{x})^{\sin{x}} > 3x $

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