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Challenge Problem #8: 3Σ(1/(√(a^3+1))≥2Σ(√(a+b))

Olinguito

Well-known member
Apr 22, 2018
251
Let $a,b,c$ be positive real numbers such that $a+b+c=2$. Prove that
$$3\left(\frac1{\sqrt{a^3+1}}+\frac1{\sqrt{b^3+1}}+\frac1{\sqrt{c^3+1}}\right)\ \geqslant\ 2\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right).$$
 

Olinguito

Well-known member
Apr 22, 2018
251
Solution:

We have
$$\frac3{a^3+1}\ =\ \frac{2-a}{a^2-a+1}+\frac1{a+1}\ \geqslant\ 2\sqrt{\dfrac{2-a}{a^3+1}}$$
putting into partial fractions and applying AM–GM (noting that all terms are positive).

Hence
$$\frac3{\sqrt{a^3+1}}\ \geqslant\ 2\sqrt{2-a}\ =\ 2\sqrt{b+c}.$$
Similarly
$$\frac3{\sqrt{b^3+1}}\ \geqslant\ 2\sqrt{c+a}$$
and
$$\frac3{\sqrt{c^3+1}}\ \geqslant\ 2\sqrt{a+b};$$
summing gives the required inequality.
 
Last edited:

kaliprasad

Well-known member
Mar 31, 2013
1,283
Solution:

We have
$$\frac3{a^3+1}\ =\ \frac{2-a}{a^2-a+1}+\frac1{a+1}\ \leqslant\ 2\sqrt{\dfrac{2-a}{a^3+1}}$$
putting into partial fractions and applying AM–GM (noting that all terms are positive).

Hence
$$\frac3{\sqrt{a^3+1}}\ \leqslant\ 2\sqrt{2-a}\ =\ 2\sqrt{b+c}.$$
Similarly
$$\frac3{\sqrt{b^3+1}}\ \leqslant\ 2\sqrt{c+a}$$
and
$$\frac3{\sqrt{c^3+1}}\ \leqslant\ 2\sqrt{a+b};$$
summing gives the required inequality.
the question says it is $>=$ but answer says it is $<=$
 

Olinguito

Well-known member
Apr 22, 2018
251
I’ve fixed the typo in my solution.