prove that $5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}$ 
prove that $$5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}$$

.A little use of calculator shows that  $\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}=5.44$.Thus the inequality is indeed true.
Generalising this result with $$f(x)=x-\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}<0$$ does not help as we see that $$8>\sqrt{8}+\sqrt[3]{8}+\sqrt[4]{8} \tag !$$
Repeated efforts of Bernoullis inequality have failed

By hand, you can compute :
$$2^2=4$$
$$1.7^3 =4.913$$
$$1.3^4=2.8561$$
So $$\sqrt{5} + \sqrt[3]{5} + \sqrt[4]{5} > \sqrt{4} + \sqrt[3]{4.913} + \sqrt[4]{2.8561}= 2 + 1.7 + 1.3 = 5$$
