A possible lower bound for $\left(1+\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}\right)^{\frac{\left(a+b+c\right)^{2}}{\left(a^{2}+b^{2}+c^{2}\right)}}$

examples-counterexamplesinequality

Problem :

Let $a,b,c>0$ then it seems we have :
$$\left(1+\frac{a}{b+c}\right)\left(1+\frac{b}{c+a}\right)\left(1+\frac{c}{a+b}\right)<\left(1+\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}\right)^{\frac{\left(a+b+c\right)^{2}}{\left(a^{2}+b^{2}+c^{2}\right)}}
$$

Some related work :

You can fin an upper bound here Prove that $(1+\frac{a^2+b^2+c^2}{ab+bc+ca})^{\frac{(a+b+c)^2}{a^2+b^2+c^2}} \leq (1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})$ with a nice solution due to user RiverLi . Unfortunetaly Bernoulli's inequality doesn't works here . I also tried to use Jensen's inequality with the logarithm but it doesn't works too . For an interesting fact let $a=b=1$ and $c\to \infty$ . To conclude on this remark we have some similarity with Nesbitt's inequality .

Edit :

Using Bernoulli's inequality we need to show :

$$1+\left(\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}\right)\frac{\left(a+b+c\right)^{2}}{\left(a^{2}+b^{2}+c^{2}\right)}>\left(1+\frac{a}{b+c}\right)\left(1+\frac{b}{c+a}\right)\left(1+\frac{c}{a+b}\right)$$

Wich seems true .

Second Edit :

For $c=0$ we have the equality :

$$\left(1+\left(\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}\right)\frac{\left(a+b+c\right)^{2}}{\left(a^{2}+b^{2}+c^{2}\right)}\right)-\left(\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{c+a}\right)+\left(1+\frac{c}{a+b}\right)\right)=0$$

Wich show the inequality in this case

Question :

How to (dis)prove it ?

Best Answer

After applying Bernoulli's inequality, note that the $a^2+b^2+c^2$ terms cancel. So we're looking at $$ 1+\frac{(a+b+c)^2}{ab + bc+ ca}> \frac{(a+b+c)^3}{(a+b)(b+c)(c+a)} $$ Multiply through by $(ab+bc+ca)$ and note that $(a+b+c)(ab+bc+ca) = (a+b)(b+c)(c+a)+abc$, giving $$ ab+bc+ca + (a+b+c)^2 > (a+b+c)^2 + \frac{abc(a+b+c)^2}{(a+b)(b+c)(c+a)} $$ Cancel terms and divide by $abc$ to get $$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} > \frac{(a+b+c)^2}{(a+b)(b+c)(c+a)} $$ At this point we can note that $$ \frac{1}{a} +\frac{1}{b}+\frac{1}{c} > \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c} = \frac{(a+b+c)^2 + ab + ac+bc}{(a+b)(b+c)(c+a)} $$ so the inequality is satisfied whenever $a,b,c>0$.

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