# Show that $7^{\sqrt {5}}>5^{\sqrt {7}}.$

Show that $7^{\sqrt {5}}>5^{\sqrt {7}}.$

I am stuck in this problem.

Any help in solving this problem will be appreciated.

• Hint: Consider the function $$f(x)=\ln((x^2)^{1/x})$$ Use calculus – Ben Grossmann May 17 '16 at 4:38
• Or Bernoulli's inequality: $(1+6)^{\sqrt{5/7}} > 1+6 \cdot \sqrt{5/7} > 1+6\cdot \frac56$ – Macavity May 17 '16 at 5:20

$7^{\sqrt {5}}>5^{\sqrt {7}}$ $\Leftrightarrow 7>5^{\sqrt{7/5}}$

$5^{\sqrt{7/5}}=5^{\sqrt{1.4}} <5^{\sqrt{1.44}}=5^{1.2}=5^{6/5}=\sqrt[5]{5^6}=\sqrt[5]{15625}<\sqrt[5]{16807}=\sqrt[5]{7^5}=7$

So, the claim is proved.

• Thanks, it was a great help. – user333900 May 17 '16 at 5:08

Showing that$$7^{\sqrt {5}}>5^{\sqrt {7}}$$ is te same as showing that $$7^{\frac{1}{\sqrt{7}}}>5^{\frac{1}{\sqrt{5}}}$$ So, now, consider the function $$f(x)=x^{\frac{1}{\sqrt{x}}}$$ After simplification, its derivative writes $$f'x)=-\frac{1}{2} x^{(\frac{1}{\sqrt{x}}-\frac{3}{2})} (\log (x)-2)$$ which cancels for $x=e^2$; so $$x<e^2\implies f'(x)>0$$ $$x=e^2\implies f'(x)=0$$ $$x>e^2\implies f'(x)<0$$ and remark that $e^2\approx 7.38906$. So, we are on a part where the function increases.

Q.E.D.

The problem is equivalent to proving that: $$\frac{\log 7}{\sqrt{7}}>\frac{\log 5}{\sqrt{5}},$$ so it is natural to consider $f(x)=\frac{\log x}{x^2}$. Its first stationary point on $\mathbb{R}^+$ occurs at $x=e^2$.

Since $e^2>7$ and $f'(1)=1>0$, $f$ is an increasing function on the interval $[5,7]$.