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# 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

Inequalities
Medium
##### Suggested Book
Inequalities by BJ Venkatachala

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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