With $a, b, c$ are positive reals and $abc=1$, prove that

Question

With $a, b, c$ are positive reals and $abc=1$, prove that:

$\sqrt[3]{\dfrac{2}{\dfrac{1+a^{2}}{2}+\dfrac{2a}{1+a}}}+\sqrt[3]{\dfrac{2}{\dfrac{1+b^{2}}{2}+\dfrac{2b}{1+b}}}+\sqrt[3]{\dfrac{2}{\dfrac{1+c^{2}}{2}+\dfrac{2c}{1+c}}}\leq3$

I don't know how to do this problem? Can you help me with this?

Answer 1

I tryed to prove it but it turned out wrong!

Try $c\rightarrow+\infty$ and $a=b$.

For $(a,b,c)=(0.01,0.01,10000)$ the left side is equal to $3.137...$.

Answer 2

For example, when $a = \frac{1}{81}, b = \frac{1}{81}, c = 6561$ the LHS is greater than 3.12.

The correct inequality appears to be $$\sqrt[3]{4}\leq\sqrt[3]{\dfrac{2}{\dfrac{1+a^{2}}{2}+\dfrac{2a}{1+a}}}+\sqrt[3]{\dfrac{2}{\dfrac{1+b^{2}}{2}+\dfrac{2b}{1+b}}}+\sqrt[3]{\dfrac{2}{\dfrac{1+c^{2}}{2}+\dfrac{2c}{1+c}}}\leq \sqrt[3]{32}$$

I wonder if Vasc's LCF Theorem can prove both sides of this...