How to prove an inequality with power

Question

How to prove that there exist a constant $c>0$ such that

$$(a^{4/3}+b^{4/3})\le c(a^2+b^2)^{1/3} $$

for all $a,b\ge 0$?

Thanks

Answer 1

Let $x= a^{1/3}$ and $y=b^{1/3}$ so we have to find $c$ such: $$(x^4+y^4)^3 \leq d(x^6+y^6)$$ where $d = c^3$ so we have for $y=1$: $$f(x):={(x^4+1)^3\over x^6+1}\leq d$$ since $f$ is unbounded such $d$ doesn't exist and so $c$ also doesn't exist.

Answer 2

By the generalized mean inequality, what you can say is that

$$ \left(\frac{a^2+b^2}{2}\right)^\frac12 \ge \left(\frac{a^{\frac43}+b^{\frac43}}{2}\right)^\frac34 \implies \left(\frac{a^{\frac43}+b^{\frac43}}{2}\right)\le \left(\frac{a^2+b^2}{2}\right)^\frac23 \\ \implies a^{\frac43}+b^{\frac43}\le 2^\frac13\left(a^2+b^2\right)^\frac23 $$

and if the RHS is $\le1$, we have

$$a^{\frac43}+b^{\frac43}\le 2^\frac13\left(a^2+b^2\right)^\frac23\le 2^\frac16\left(a^2+b^2\right)^\frac13$$