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

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The discussion revolves around proving the inequality involving positive real numbers a, b, and c, constrained by the condition a + b + c = 2. The main inequality to be proven is that 3 times the sum of the reciprocals of the square roots of (a^3 + 1) is greater than or equal to 2 times the sum of the square roots of the pairs (a+b), (b+c), and (c+a). A participant mentions correcting a typo in their solution, indicating ongoing refinement of the proof. The focus remains on demonstrating the validity of the mathematical statement through appropriate methods. The conversation highlights the importance of clarity and accuracy in mathematical proofs.
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Let $a,b,c$ be positive real numbers such that $a+b+c=2$. Prove that
$$3\left(\frac1{\sqrt{a^3+1}}+\frac1{\sqrt{b^3+1}}+\frac1{\sqrt{c^3+1}}\right)\ \geqslant\ 2\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right).$$
 
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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:
Olinguito said:
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 $<=$
 
I’ve fixed the typo in my solution.
 
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