Just use holder on each term $3+9+27+81=120$

For completeness, details are \[ (x^n+y^n+z^n)\underbrace{(1+1+1)\cdots(1+1+1)}_{(n-1)\text{ times}}\ge (x+y+z)^n, \]yielding that the LHS is at most $3^1+3^2+3^3+3^4 = 120$.

Since the function $f(x)=x^n$ convex if $n\geq 2$.So by $Jensen$'s inequality we have $\frac{f(a)+f(b)+f(c)}{3}\geq f(\frac{a+b+c}{3})$.Then result follows.

parmenides51 wrote: Let $a, b, c, d$ be four positive real numbers. Prove that $$\frac{(a + b + c)^2}{a^2+b^2+c^2}+\frac{(b + c + d)^3}{b^3+c^3+d^3}+\frac{(c+d+a)^4}{c^4+d^4+a^4}+\frac{(d+a+b)^5}{d^5+a^5+b^5}\le 120$$ h

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