Prove that $\sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ac+b} \ge 1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

Question

It is given that $a+b+c=1$ and $a,b,c$ are positive real numbers

Then I reduced the inequality to

$\sqrt{(1-a)(1-b)}+\sqrt{(1-b)(1-c)}+\sqrt{(1-c)(1-a)} \ge 1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}$.

I know that

$\sqrt{(1-a)(1-b)} \ge \sqrt{ab}$. Similarly we can deduce other too.By this we get

$\sqrt{(1-a)(1-b)}+\sqrt{(1-b)(1-c)}+\sqrt{(1-c)(1-a)} \ge \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

I don't know how they got 1 on rhs.Any help??

You have $\sqrt{ab+c}\geq \sqrt{ab}+c$, and the likes. This can be proved with AM-GM, Cauchy-Schwarz, or simple expansion. So, $$\sum \sqrt{ab+c}\geq \sum\sqrt{ab}+\sum c=\sum\sqrt{ab}+1.$$ — Batominovski, Mar 21 '20 at 14:07
By AM-GM, $a+b\ge 2\sqrt ab$. So $ab+c=ab+(a+b)c+c^2\ge ab+2\sqrt{ab}c+c^2$. — Batominovski, Mar 21 '20 at 14:29
I think it does not prove your result.can you make it more clear?? — Mathematical Curiosity, Mar 21 '20 at 14:44
How much clearer do you want? I just proved that $$\sqrt{ab+c}^2=ab+c\ge ab+2\sqrt{ab}c+c^2=(\sqrt{ab}+c)^2.$$ How does it not prove my result? I think, perhaps you should start with simpler exercises for AM-GM to better see how to use AM-GM. — Batominovski, Mar 21 '20 at 14:47
@WETutorialSchool I arrived finally to the equation on my own which is essentially same as your comments and added my answer. (Deleted my previous answer though) but I see that it has been already answered — Kartoos, Mar 21 '20 at 15:30
Does this answer your question? $a,b,c>0 \text{ s.t. }a+b+c=1 \implies \sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ca+b} \ge 1+ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}$ — Martin R, Mar 21 '20 at 17:37

Kartoos · Answer 1 · 2020-03-21T15:34:56.567

4

By AM-GM inequality $$\frac{(a+b)}{2} \ge \sqrt{ab}$$

$$ a+b \ge 2\sqrt{ab} $$ $$ ca+cb \ge 2c\sqrt{ab} $$ $$ ca+cb+c^2+ab \ge c^2+ab+ 2c\sqrt{ab} $$ $$ c(a+b+c)+ab \ge c^2+ab+ 2c\sqrt{ab} $$ $$ c+ab \ge \left(c+\sqrt{ab}\right)^2 $$ $$ \sqrt{c+ab} \ge c+\sqrt{ab} $$ $$ \implies \sum_{cyc}\sqrt{c+ab} \ge \sum_{cyc}c+\sum_{cyc}\sqrt{ab} $$ $$ \implies \sum_{cyc}\sqrt{ab+c} \ge 1+\sum_{cyc}\sqrt{ab} $$ which is same as the given equation, hence proved

edited Mar 21 '20 at 15:34

answered Mar 21 '20 at 15:25

Kartoos

190

I hope I did not do any mistake this time.. – Kartoos Mar 21 '20 at 15:25
Seems good to me – Maximilian Janisch Mar 21 '20 at 15:45

Michael Rozenberg · Accepted Answer · 2020-03-21T14:15:34.197

By C-S $$\sum_{cyc}\sqrt{a+bc}=\sum_{cyc}\sqrt{a(a+b+c)+bc}=\sum_{cyc}\sqrt{(a+b)(a+c)}=$$ $$=\sqrt{\sum_{cyc}((a+b)(a+c)+2(a+b)\sqrt{(a+c)(b+c)}}\geq$$ $$\geq\sqrt{\sum_{cyc}((a+b)(a+c)+2(a+b)(\sqrt{ab}+c)}=$$ $$=\sqrt{\sum_{cyc}(a^2+7ab+2\sqrt{a^3b}+2\sqrt{a^3c})}=$$ $$=\sqrt{(a+b+c)^2+\sum_{cyc}(2\sqrt{a^3b}+2\sqrt{a^3c}+5ab)}.$$ Thus, it's enough to prove that: $$(a+b+c)^2+\sum_{cyc}(2\sqrt{a^3b}+2\sqrt{a^3c}+5ab)\geq$$ $$\geq(a+b+c)^2+2(a+b+c)\sum_{cyc}\sqrt{ab}+\sum_{cyc}(ab+2a\sqrt{bc})$$ or $$\sum_{cyc}(4ab-4c\sqrt{ab})\geq0$$ or $$\sum_{cyc}c(\sqrt{a}-\sqrt{b})^2\geq0$$ and we are done!

The equality $(a+c)(b+c)\geq(\sqrt{ab}+c)^2$ we can prove also by AM-GM: $$(a+c)(b+c)-(\sqrt{ab}+c)^2=ab+ac+bc+c^2-ab-2c\sqrt{ab}-c^2=$$ $$=ac+bc-2c\sqrt{bc}\geq2\sqrt{ac\cdot bc}-2c\sqrt{bc}=0,$$ which gives much more easier solution: $$\sum_{cyc}\sqrt{a+bc}=\sum_{cyc}\sqrt{(a+b)(a+c)}\geq\sum_{cyc}(a+\sqrt{bc})=1+\sum_{cyc}\sqrt{ab}.$$

thanks for your answer but can you provide an answer using on am gm because this problem is meant to be solved by am gm and I am curious how it can be done — Mathematical Curiosity, Mar 21 '20 at 14:01

Prove that $\sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ac+b} \ge 1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

2 Answers2