For $a,b,c>0$. Prove that $$\sum \sqrt{{\frac {2{a}^{2}b}{a+c}}} \leqq a+b+c \,\,—–(1)$$
My solution$:$
By C-S, we need to prove: $$(\sum ab) \cdot (\sum \frac{2a}{a+c}) \leqq (a+b+c)^2\, (\ast)$$
+) First way to prove $(\ast)$$:$
Let $c=\min\{a,b,c\}$$,$ it is equivalent to$:$
$$2\,c \left( a-b \right) ^{2} \left( b+a \right) ^{2}+ \left( b-c \right) \left( a-c \right) \left\{ {a}^{3}+2\,c{a}^{2}+a{b}^{2}+2\,a bc+{c}^{2}a+{c}^{2}b+b \left( a-b \right) ^{2} \right\} \geqq 0$$
+) And the second$:$ We have:
$$\text{LHS} = (ab+bc+ca)(6 – \sum \frac{2c}{a+c})$$
$$\leqq (ab+bc+ca) \cdot \Big[6- \frac{2(a+b+c)^2}{(ab+bc+ca) +(a^2+b^2+c^2)}\Big]$$
$$=\left( a+b+c \right) ^{2}-{\frac { \left( {a}^{2}-ab-ca+{b}^{2}-bc+{c }^{2} \right) \left( {a}^{2}+{b}^{2}+{c}^{2} \right) }{{a}^{2}+ab+ca+ {b}^{2}+bc+{c}^{2}}} \leqq (a+b+c)^2$$
PS: Actually the original inequality is$:$
For $a,b,c>0$ and $abc=1$$.$ Prove$:$
$$\sum \sqrt{\frac{ab}{bc^2 +1}} \leqq \frac{a+b+c}{\sqrt{2}}$$
I found $(1)$ when I try to take homogeneous for the original inequality.
What are some other proofs for $(1)$ or the original inequality?
Best Answer
By C-S $$\sum_{cyc}\sqrt{\frac{a^2b}{a+c}}\leq\sqrt{\sum_{cyc}\frac{ab}{(a+c)(b+c)}\sum_{cyc}a(b+c)}.$$ Thus, it's enough to prove that: $$\sum_{cyc}\frac{ab}{(a+c)(b+c)}\sum_{cyc}a(b+c)\leq\frac{(a+b+c)^2}{2}$$ or $$(a+b+c)^2\prod_{cyc}(a+b)\geq4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab,$$ which is obviously true by $uvw$, but there is a nice solution by SOS.
Indeed, let $a\geq b\geq c$.
Thus, $$(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab=$$ $$=\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2-2a^3bc+2a^2b^2c^2)=$$ $$=\sum_{cyc}(a^4b-a^3b^2-a^2b^3+ab^4)-abc\sum_{cyc}(a^2-2ab+b^2)=$$ $$=\sum_{cyc}(a-b)^2ab(a+b-c)\geq(a-c)^2ac(a+c-b)+(b-c)^2bc(b+c-a)\geq$$ $$\geq(b-c)^2ac(a-b)+(b-c)^2bc(b-a)=(b-c)^2(a-b)^2c\geq0.$$
A full expanding we can make by the following way: $$(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab=$$ $$=\sum_{cyc}(a^2+2ab)\sum\limits_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab=$$ $$=\sum_{cyc}(a^4b+a^4c+a^3b^2+a^3c^2+2a^2b^2c+2a^3bc+2a^3b^2+2a^3c^2+4a^3bc+4a^2b^2c+4a^2b^2c)-$$ $$-4\sum_{cyc}(a^3b^2+a^3c^2+2a^3bc+2a^2b^2c)=$$ $$=\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2-2a^3bc+2a^2b^2c).$$