Without a calculator, prove: $7^{1/2} + 7^{1/3} + 7^{1/4} < 7$ [solved] I need to prove
$$7^{1/2} + 7^{1/3} + 7^{1/4} < 7$$
I have reached the stage $$1 + 7^{-2/12} + 7^{-3/12} < 7^{1/2}$$, but don’t know if I’m on the right track or what to do next. Thanks
Thanks for the help everyone, much appreciated.
 A: We have $\sqrt{7}<3$, $\sqrt[3]{7}<2$ and $\sqrt[4]{7}<2$. Summing up
$$\sqrt{7}+\sqrt[3]{7}+\sqrt[4]{7}<3+2+2=7$$
A: Hint: Consider the numbers $9,8,16$. Think of how you can use them to construct an inequality on the LHS. (If you don't get the solution in one-two lines, you're doing something wrong/long-winded)
A: $$\sqrt[3]7<2,$$
$$\sqrt7<2.65$$ because $$26\cdot27=676+26=702$$ and from here $$2.65^2=7.0225$$ and 
$$\sqrt[4]7<\sqrt{2.65}<\sqrt{2.89}=1.7.$$
Id est, $$\sqrt7+\sqrt[3]7+\sqrt[4]7<2+2.65+1.7<7.$$
A: One way (if you have learned calculus) is to consider the function $$f(x) = x-x^{1/2}-x^{1/3}-x^{1/4}$$
We can numerically find zeroes close to $0$ and $6$. Then we can take a look at the derivative:
$$f'(x) = 1-\frac{1}{2}x^{-1/2}-\frac{1}{3}x^{-2/3}-\frac{1}{4}x^{-3/4}$$
The $1$ is constant term and everything else is strictly moving towards $0$ with increasing $x$. So whenever $x$ becomes large enough the $1$ is sure to "win" over the variable terms.
Should be easily finishable from here.
This can be used to prove the inequality for all large enough $x$, so it also works for 8,9,10 et.c.
