Solution March 20, 2007

Problem

$a,\,b,\,c,\,d$ are positive real numbers satisfying the following condition: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=4$

Prove that: $\sqrt[3]{\frac{a^{3}+b^{3}}{2}}+\sqrt[3]{\frac{b^{3}+c^{3}}{2}}+\sqrt[3]{\frac{c^{3}+d^{3}}{2}}+\sqrt[3]{\frac{d^{3}+a^{3}}{2}}\leq 2(a+b+c+d)-4$

Solution

For brevity, define

$L := \sqrt[3]{\frac{a^{3}+b^{3}}{2}}+\sqrt[3]{\frac{b^{3}+c^{3}}{2}}+\sqrt[3]{\frac{c^{3}+d^{3}}{2}}+\sqrt[3]{\frac{d^{3}+a^{3}}{2}}$

We will show that

$L\leq 2(a+b+c+d)-\frac{16}{a^{-1} + b^{-1} + c^{-1} + d^{-1}}\qquad(1)$

whenever $a,\,b,\,c,\,d> 0$ .

Lemma: (1) is true for all $a,\,b,\,c,\,d> 0$ if and only if

$\sqrt[3]{\frac{u^{3}+v^{3}}{2}}\leq u+v-\frac{2}{u^{-1} + v^{-1}}\qquad(2)$

holds for all $u,\,v>0$

Proof: It's easy to see that (2) follows from (1): just set $a=u,\,b=v,\,c=u,\,d=v$ .

So assume that (2) is true for all $u,\,v>0$ and let $a,\,b,\,c,\,d>0$ . Sum (2) for $(u,v)\in\{(a,b),\,(b,c),\,(c,d),\,(d,a)\}$ . We obtain

$L\leq 2(a+b+c+d)-\frac{2}{a^{-1} + b^{-1}}-\frac{2}{b^{-1} + c^{-1}}-\frac{2}{c^{-1} + d^{-1}}-\frac{2}{d^{-1} + a^{-1}}\qquad(3)$

Now by the arithmetic-harmonic means inequality (http://planetmath.org/encyclopedia/ArithmeticGeometricMeansInequality.html),

$\frac14\left(\frac{2}{a^{-1} + b^{-1}}+\frac{2}{b^{-1} + c^{-1}}+\frac{2}{c^{-1} + d^{-1}}+\frac{2}{d^{-1} + a^{-1}}\right)\geq \frac{4}{\frac{a^{-1} + b^{-1}}2+\frac{b^{-1} + c^{-1}}2+\frac{c^{-1} + d^{-1}}2+\frac{d^{-1} + a^{-1}}2}$

and (1) follows from that and (3), completing the proof of the lemma. $\square$

All that's left to show is (2). For this purpose, let $x:=\frac{uv}{(u+v)^2}$ . It's easy to see (http://planetmath.org/encyclopedia/ArithmeticGeometricMeansInequality.html) that $x\leq \frac14$ .

Divide (2) by $(u+v)\,$ , raise it to the 3rd power and simplify:

$\frac12 - \frac{3uv}{2(u+v)^2} \leq (1 - 2x)^3$

$1 - 3x \leq 2 - 12x + 24x^2 - 16x^3$

$0\leq (1-4x)^2(1-x)$

The latter inequality is true for all $x\leq 1$ . This completes the proof of (2), and by our Lemma, proves (1). $\square$

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