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