Understand the problem

Let $a,b,c,d$ be positive real numbers such that $abcd=1$.Find with proof that $x=3 $ is the minimal value for which the following inequality holds:
\[a^x+b^x+c^x+d^x\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\]

Source of the problem

Albania IMO TST 2013

Topic
Inequalities
Difficulty Level
Medium
Suggested Book
Inequalities by BJ Venkatachala

Start with hints

Do you really need a hint? Try it first!

Choose specific values of a,b,c,d to show that no value of x less than 3 works.
Choosing a=b=c=t, the inequality becomes 3t^x+\frac{1}{t^{3x}}\ge \frac{3}{t}+t^3. Note that, as t\to\infty, the LHS grows as O(t^x) and the RHS grows as O(t^3). For the LHS to be always greater than the RHS, it should be of a higher order. Hence, $x\ge 3$. Now show that the inequality is true for x=3.
Note that the RHS can be rewritten as abc+bcd+cda+bda. This reminds us of the AM-GM inequality.
Indeed, abc+bcd+cda+bda\le \frac{a^3+b^3+c^3}{3}+\frac{d^3+b^3+c^3}{3}+\frac{a^3+d^3+c^3}{3}+\frac{a^3+b^3+d^3}{3}=a^3+b^3+c^3+d^3.

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