$2\cdot5^x-7^x-4^x>0$ for $-1\le x<0$

Question

Show that $2\cdot5^x-7^x-4^x>0$ for $-1\le x<0$.

I tried by differentiation but I found it useless as expression become more complicated. Also tried by Jensen's inequality but did not succeed.

Answer 1

Let $f(x)=2\cdot 5^x-7^x-4^x$ and let $$g(x)=f(x)/5^x=2-(7/5)^x-(4/5)^x.$$ Then $$g'(x)=\log(5/4)(4/5)^x-\log(7/5) (7/5)^x.$$ Note that as $x$ increases from $-1$ to $0$, $(4/5)^x$ decreases and $(7/5)^x$ increases, so $g'(x)$ is strictly decreasing on $[-1,0]$. In particular, $g$ cannot have a local minimum on $(-1,0)$, and so its minimum on $[-1,0]$ is achieved only at one or both of the endpoints. We can compute that $g(-1)>0$ and $g(0)=0$, and so $x=0$ is the unique minimum of $g$ on $[-1,0]$. Thus $g(x)>0$ on all of $[-1,0)$, and so $f(x)>0$ on $[-1,0)$.

Answer 2

Note that $p^x$ is convex for $p>0$, so on $[a,b]$ it doesn't lie above a line passing through endpoints $(a,p^a) \text{ and } (b,p^b)$. Divide the initial inequality by $5^x$ and bound from above by linear pieces: $$\Big(\frac 4 5\Big)^x\le 1+x\big(1-\frac 5 4\big)$$ $$\Big(\frac 7 5\Big)^x\le 1+x\big(1-\frac 5 7\big)$$ on $[-1,0]$. Adding up we get $$\Big(\frac 4 5\Big)^x+\Big(\frac 7 5\Big)^x\le 2+\frac 1 {28}x<2$$ on $[-1,0)$. Q.E.D.

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Alternatively, building on Eric Wofsey's idea, note that LHS of the last inequality is a sum of two strictly convex functions, so its maximum is achieved at the endpoints which can be checked. This saves you taking and analyzing a derivative directly.