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

Question

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

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??

Answer 1

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: $$\sum_{cyc}\sqrt{a+bc}=\sum_{cyc}\sqrt{(a+b)(a+c)}\geq\sum_{cyc}(a+\sqrt{bc})=1+\sum_{cyc}\sqrt{ab}.$$

Answer 2

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