If $ { a\over a+1} + { b\over b+1 } + { c\over c+1 } = 1 $ prove $ abc \le 1/8 $

Question

If $ a,b,c $ are positive real numbers and $ { a\over a+1} + {b\over b+1} + {c\over c+1} = 1 $ then prove that $ abc \le {1\over 8} $

Could I get some help with this?

What I have tried-

Making the denominators equal and adding to get another equation of the form $ {N\over M}= 1 $ then multiplying by $ M $ both the sides and simplifying to get $ 2abc + ab + bc + ac = 1 $ , from which we have, $ abc \lt 1/2 $

But which obviously isn't enough to prove that $ abc \le 1/8 $

I also tried using Jensen's inequality which states that $ f(E(X)) \ge E(f(X)) $ when $ f $ is a concave function and $ f(E(X)) \le E(f(X)) $ when $ f $ is a convex function, and where $ E(X) $ denotes the expectation of X ,

but that didn't really help.

Answer 1

What you have ( $2abc+ab+bc+ca=1$ and $a,b,c>0$ ) does actually imply $abc\leq 1/8$. This is because, by AM-GM, $$ab+bc+ca\geq3\sqrt[3]{a^2b^2c^2}$$and so, writing $x=abc$, we have $1\geq 2x+3x^{2/3}.$ The RHS is clearly increasing in $x$, and we have equality when $x=1/8$, so any $x>1/8$ violates this inequality.

Answer 2

The AM-GM inequality on $2abc$, $ab$, $ac$, $bc$ gives:

$$1 = 2abc+ab+bc+ca \geq 4 \sqrt[4]{2a^3b^3c^3}$$

So $1 \geq 2^{9/4} (abc)^{3/4}$. This gives $2^{-9/4} \geq (abc)^{3/4}$, so $2^{-3} = 2^{-9/4 \cdot 4/3} \geq abc$. So $abc \leq \frac18$.

Answer 3

If you multiply both sides by $(a+1)(b+1)(c+1)$ you'll obtain $$3abc+2ab+2ac+2bc+a+b+c=(a+1)(b+1)(c+1),$$

which simplifies to $$3abc+2ab+2ac+2bc+a+b+c=1 + a + b + a b + c + a c + b c + a b c,$$

rearranging to $$abc=\frac{1-(ab+ac+bc)}{2}.$$

Using AM-GM you get $$1-2abc=ab+ac+bc\geq3\sqrt[3]{a^2b^2c^2}.$$ Now set $u=abc$ and you have $$u=\frac{1-\sqrt[3]{u^2}}{2}$$ which leads to $$1\geq2u+3u^{2/3}.$$ Now you see that $u=abc>1/8$ violates the inequality.

Answer 4

Substitute \begin{eqnarray*} a =\frac{x}{y+z} \\ b=\frac{y}{z+x} \\ c= \frac{z}{x+y} \end{eqnarray*} This satisfies the constraint. Now the following quantity is clearly positive \begin{eqnarray*} x(y-z)^2+y(z-x)^2+z(x-y)^2 \geq 0 \\ (x+y)(y+z)(z+x) \geq 8xyz \\ \end{eqnarray*} Rearrange a bit & substitute back in & the result follows.

Answer 5

By AM-GM:$$\frac{1}{a+1}=1-\frac{a}{a+1}=\frac{b}{b+1}+\frac{c}{c+1}\geq2\sqrt{\frac{bc}{(b+1)(c+1)}}.$$ By the permutation symmetry, the inequality is invariant with respect to the cyclic permutation of $(a,b,c)$. Thus, $$\frac{1}{(a+1)(b+1)(c+1)}\geq\frac{8abc}{(a+1)(b+1)(c+1)}$$ and we are done!