# Show that $a+b+c \leq 2 + 2abc$ for $0 \leq a,b,c \leq 1$.

I want to show that $$a+b+c \leq 2 + 2abc$$ for $$0 \leq a,b,c \leq 1$$.

I think this inequality is true, but I don't know how to show it properly. I have tried substituting $$a=b=c=t$$, which gives $$3t \leq 2+2t^3$$, which seems to be true.

I tried $$a=b=(1-t)$$, so we have $$2-2t+c \leq 2 + 2(1-t)^2 c$$, which I work out to be true also.

I can show that the easier inequality $$a+b+c \leq 2 + 2ab$$ (assuming that $$a).

$$2+2abc-(a+b+c)=(1-a)(1-b)+(1-ab)(1-c)+abc\ge 0.$$ Moreover, a better inequality would be : $$a+b+c\le 2+abc,$$ and you can prove by induction that :

$$\sum_{k=1}^n a_k\le n-1+\prod_{k=1}^n a_k,$$ where $$0\le a_k\le 1$$.

Direct proof - $$n-1+\prod_{k=1}^n a_k-\sum_{k=1}^n a_k=(1-a_1)(1-a_2)+(1-a_1a_2)(1-a_3)+...+(1-a_1...a_{n-1})(1-a_n)\ge 0.$$

This result is often used to prove other inequalities, like :

Problem 1 - If $$1\le a_k\le 1$$, then : $$\sum_{k=1}^n \frac{1}{1+a_k}\le n-1+\frac{1}{1+\displaystyle\prod_{k=1}^n a_k}.$$

And even results about the roots of polynomials :

Problem 2 - If $$P=X^n+a_1X^{n-1}+...a_n$$ is a complex polynomial with roots $$z_1,...,z_n$$, then : $$\sum_{i=1}^n |z_i|^2 \le n-1+\sum_{i=1}^n |a_i|^2.$$

• Great answer, thank you! – eatfood Aug 9 '19 at 10:18

It's a linear inequality of any variables, which says that it's enough to check: $$\{a,b,c\}\subset\{0,1\},$$ which gives that our inequality is true.

Easy to see that even $$a+b+c\leq2 +abc$$ is true with the same conditions.

• But RHS of the inequality is not linear: $2+2abc$. – Z Ahmed Aug 9 '19 at 10:02
• No, it's linear for every variable if the other variables are fixed. – richrow Aug 9 '19 at 10:33