Prove the inequalitiy

Question

If $a, b, c$ are positive reals such that: $\frac a {b + c + 1} + \frac b {a + c + 1} + \frac c {b + a + 1} \le 1$ then:

$\frac 1 {b + c + 1} + \frac 1 {a + c + 1} + \frac 1 {b + a + 1} \ge 1$

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I tried using inequality of arithmetic and harmonic means, to no avail.

Any help is appreciated.

Answer 1

We'll write the condition in the following form: $$a+b+c+1\geq a^3+b^3+c^3+abc+\sum\limits_{cyc}(a^2-ab)$$

and since $\sum\limits_{cyc}(a^2-ab)\geq0$ we obtain $$a+b+c+1\geq a^3+b^3+c^3+abc$$ In another hand, we need to prove that $$2(a+b+c+1)\geq(a+b)(a+c)(b+c)$$ Thus, it remains to prove that $$2(a^3+b^3+c^3+abc)\geq(a+b)(a+c)(b+c)$$ which is $\sum\limits_{cyc}(a+b)(a-b)^2\geq0$. Done!

Answer 2

.WLOG, $a <b <c$, then $\frac{1}{b+c} < \frac{1}{a+c} < \frac{1}{a+b}$

Note that $$\frac{a}{b+c+1} + \frac{b}{a+c+1} + \frac{c}{a+b+1} = (a+b+c+1)\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) - 3$$

Hence, $$ (a+b+c+1)\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) \leq 4 $$ Further, by the harmonic mean-arithmetic mean inequality, $$ \bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) \geq \dfrac{9}{2(a+b+c)+3} $$ Now, let $x=\bigg(\frac{1}{b+c+1} + \frac{1}{a+c+1} + \frac{1}{a+b+1}\bigg) $ and $y = a+b+c$. Then, we have: $2yx+3x \geq 9$ and $4 \geq yx + x$. Multiplying by $2$ and changing the sign, $-2yx-2x \geq 8$. Adding ,we get $x \geq 1$. Hence the result.