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Show that if $a,b,c$ are positive real numbers such that $S_2 := ab+bc+ca = 3$, then

$$ \cfrac{1}{a+b+2} + \cfrac{1}{a+c+2} + \cfrac{1}{b+c+2} \leq \cfrac{13 \cdot S_1 + 27}{16 \cdot S_1 + 40}$$

where $S_1 = a+b+c$

My road so far

Well, my idea so far was to find some expression "more obviously" between the two sides of that inequality:

$$ \cfrac{1}{a+b+2} + \cfrac{1}{a+c+2} + \cfrac{1}{b+c+2} \leq f(a,b,c) \leq \cfrac{13 \cdot S_1 + 27}{16 \cdot S_1 + 40}$$

An obvious guess would be just set $a=b=c=1$, resuilting $f(1,1,1) = \frac{3}{4}$.

The rightmost inequality is not so hard: it just implies $S_1 \geq3$, a classical result.

But I am stuck in the leftmost inequality...

Anderson Torres
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4 Answers4

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Because $1-\frac{2}{a+b+c}=\frac{a+b}{a+b+2},$ so inequality equivalent to $$\sum \frac{a+b}{a+b+2} \geqslant \frac{11(a+b+c)+33}{8(a+b+c)+20}.$$ From known inequality (here) $$\sum \sqrt{(a+b)(b+c)} \geqslant a+b+c+\sqrt{3(ab+bc+ca)},$$ and the Cauchy-Schwarz inequality, we get $$\sum \sqrt{(a+b)(b+c)} \geqslant \frac{\displaystyle \left(\sum \sqrt{a+b}\right)^2}{\displaystyle \sum (a+b+2)} = \frac{\displaystyle a+b+c+\sum \sqrt{(a+b)(b+c)}}{a+b+c+3}$$ $$ \geqslant \frac{2(a+b+c)+\sqrt{3(ab+bc+ca)}}{a+b+c+3}.$$ Thefore, we will show that $$\frac{2(a+b+c)+\sqrt{3(ab+bc+ca)}}{a+b+c+3} \geqslant \frac{11(a+b+c)+33}{8(a+b+c)+20}. \tag 1$$ Let $x = a+b+c, \; y = \sqrt{3(ab+bc+ca)}$ then $ x\geqslant y,$ inequality (1) become $$\frac{2x+y}{x+y} \geqslant \frac{33(x+y)}{4(6x+5y)},$$ equivalent to $$4(6x+5y)(2x+y) \geqslant 33(x+y)^2,$$ or $$(15x+13y)(x-y) \geqslant 0.$$ Which is true. The proof is completed.

nguyenhuyenag
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Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, since $$\prod_{cyc}(a+b)=9uv^2-w^3$$ and $\sum_{cyc}(a+b+2)(a+c+2)$ and $\frac{13(a+b+c)+27}{16(a+b+c)+40}>0$ are not depended on $w^3$,

we see that our inequality is equivalent to $f(w^3)\leq0,$ where $f$ increases.

Thus, it's enough to prove our inequality for a maximal value of $w^3$,

which by $uvw$ happens for equality case of two variables.

Let $b=a$.

Thus, $c=\frac{3-a^2}{2a}$ and we obtain: $$(a-1)^2(13a^2+24a+15)\geq0,$$ which is obvious.

Michael Rozenberg
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The pqr method

Let $p = a + b + c, q = ab + bc + ca = 3$ and $r = abc$. We need to prove that $$\frac{p^2+8p+q+12}{2p^2+pq+8p+2q-r+8} \le \frac{13p+27}{16p+40}.$$ Since $p^2 \ge 3q$, we have $p \ge 3$. Since $q^2 \ge 3pr$, it suffices to prove that $$\frac{p^2+8p+q+12}{2p^2+pq+8p+2q-\frac{q^2}{3p}+8} \le \frac{13p+27}{16p+40}$$ or $$\frac{(p-3)(10p^2+29p+9)}{8(2p^2+5p-1)(2p+5)}\ge 0.$$ We are done.

River Li
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Now I will post my solution after following some ideas above.

Let $a+b+c=S_1, ab+bc+ca=S_2, abc=S3$.

We want to show that

$$\cfrac{1}{a+b+2} + \cfrac{1}{a+c+2} + \cfrac{1}{b+c+2} \leq \cfrac{3}{4} \leq \cfrac{13 \cdot S_1 + 27}{16 \cdot S_1 + 40}$$

First the hard part:

$$\cfrac{1}{4} - \cfrac{1}{a+b+2} = \cfrac{a+b+2-2}{4(a+b+2)} = \cfrac{1}{4} \cdot \cfrac{(a+b-2)}{(a+b+2)}$$

We want to show that $$ \cfrac{(a+b-2)}{(a+b+2)} + \cfrac{(a+c-2)}{(a+c+2)} + \cfrac{(a+b-2)}{(a+b+2)} \geq 0$$

Or, opening:

$$ (a+b-2)(a+c+2)(b+c+2) + (a+b+2)(a+c-2)(b+c+2) + (a+b+2)(a+c+2)(b+c-2) \geq 0$$

But $$(a+b-2)(b+c+2)(c+a+2) = (a^2b+a^2c+ab^2+ac^2+b^2c+bc^2) + (2a^2+2b^2-2c^2) + (2ab+2ac+2bc) + 2abc -8c-8$$

So, adding them we have

$$(a+b-2)(a+c+2)(b+c+2) + (a+b+2)(a+c-2)(b+c+2) + (a+b+2)(a+c+2)(b+c-2) = 3(a^2b+a^2c+ab^2+ac^2+b^2c+bc^2) + 2(a^2+b^2+c^2) + 6(ab+ac+bc) + 6abc -8(a+b+c)-24 $$

But $(a^2b+a^2c+ab^2+ac^2+b^2c+bc^2) = (a^2b+ab^2+abc) + (a^2c+ac^2+abc) + (b^2c+bc^2+abc) -3abc = S_1S_2 - 3S_3$

And $(a^2+b^2+c^2) = S_1^2-2S_2$

Substituting and remembering $S_2=3$ we have

$$(a+b-2)(a+c+2)(b+c+2) + (a+b+2)(a+c-2)(b+c+2) + (a+b+2)(a+c+2)(b+c-2) = 3(S_1S_2 - 3S_3) + 2(S_1^2-2S_2) + 6S_2 + 6 -8S_1 - 24 = 3S_1S_2 - 9S_3 + 2S_1^2-4S_2 + 6S_2 + 6S_3 - 8S_1 - 24 = 9S_1 - 9S_3 + 2S_1^2-12 + 18 + 6S_3 - 8S_1 - 24 = 2S_1^2 + S_1 - 3S_3 - 18 $$

By the Means Inequality, we have $$\cfrac{S_1}{3} \geq \sqrt{\cfrac{S_1}{3}} \geq \sqrt[3]{S_3}$$. Substituting $S_2$, we have $S_1 \geq 3$ and $S_3 \geq 1$.

So our expression is at least $2 \cdot 3^2 + 3 - 3 \cdot 1 - 18 = 0$, as we wish!

The rightmost inequality is just an equivalent form of $S_1 \geq 3$.

Anderson Torres
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