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#### Olinguito

##### Well-known member

- Apr 22, 2018

- 251

$$3\left(\frac1{\sqrt{a^3+1}}+\frac1{\sqrt{b^3+1}}+\frac1{\sqrt{c^3+1}}\right)\ \geqslant\ 2\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right).$$

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- Thread starter
- #1

- Apr 22, 2018

- 251

$$3\left(\frac1{\sqrt{a^3+1}}+\frac1{\sqrt{b^3+1}}+\frac1{\sqrt{c^3+1}}\right)\ \geqslant\ 2\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right).$$

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- 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.

$$\frac3{a^3+1}\ =\ \frac{2-a}{a^2-a+1}+\frac1{a+1}\ \geqslant\ 2\sqrt{\dfrac{2-a}{a^3+1}}$$

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:

- 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 $<=$

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- #4

- Apr 22, 2018

- 251

I’ve fixed the typo in my solution.