Given three positive numbers $a,b,c$. Prove that $\sum\limits_{sym}\frac{a+b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$ .

Question

(A problem due to Mr. Le Khanh Sy). Given three positive numbers $a, b, c$. Prove that $$\sum\limits_{sym}\frac{a+ b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$$

I'm eagerly interested in learning one method which assumes $c\not\equiv {\rm mid}(\!a, b, c\!)$. But if $c\equiv {\rm mid}(\!a, b, c\!)$: $$2\sqrt{(\!\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab}\!)(\!a+ b+ c\!)}\leqq c(\!\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab}\!)+ \frac{a+ b+ c}{c}= \frac{a+ b}{c}+ \frac{a}{b}+ \frac{b}{a}+ 1+ \frac{c^{2}}{ab}$$ We need to prove $$\begin{align} \frac{a+ b}{c}+ \frac{a}{b}+ \frac{b}{a}+ 1+ \frac{c^{2}}{ab}\leqq \frac{a+ b}{c}+ \frac{b+ c}{a} & + \frac{c+ a}{b}\Leftrightarrow 1+ \frac{c^{2}}{ab}\leqq \frac{c}{a}+ \frac{c}{b}\Leftrightarrow \\ & \Leftrightarrow (\frac{c}{a}- 1)(\frac{c}{b}- 1)\leqq 0\Leftrightarrow \frac{(c- a)(c- b)}{ab}\leqq 0 \end{align}$$ Who can teach me what would we do if $c\not\equiv {\rm mid}(\!a, b, c\!)$ ? I am goin' to set a bounty, thank u so much

Answer 1

After squaring of the both sides we need to prove that $$\sum_{sym}(a^4b^2-a^4bc+a^3b^3-2a^3b^2c+a^2b^2c^2)\geq0,$$ which is true by Muirhead and Schur.

Answer 2

As I have written, the idea of proof taking $c=mid(a,b,c)$ is just a matter of notation. Without loss of generality (WLOG), You can take the other mid one if $c$ is not the one. Note that there is always something in the middle. Even if $a=b=c$ you can take any of them.

Answer 3

Let $c\neq {\rm mid}\{a,b,c\}$ and $c'={\rm mid}\{a,b,c\}.$

Now, let another variables be $a'$ and $b'$.

Thus, since our inequality does not depend on any permutations of $a'$, $b'$ and $c'$, we need to prove that: $$\frac{a'+b'}{c'}+\frac{b'+c'}{a'}+\frac{c'+a'}{b'}\geq2\sqrt{(a'+b'+c')\left(\frac{a'}{b'c'}+\frac{b'}{c'a'}+\frac{c'}{a'b'}\right)},$$ where $a'$, $b'$ and $c'$ are positives such that $c'={\rm mid}\{a',b',c'\},$ which you made already.

Answer 4

By the Vornicu Schur, we have $$\left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}} \right)\left( {\frac{a}{c} + \frac{c}{b} + \frac{b}{a}} \right) - \left( {a + b + c} \right)\left( {\frac{a}{{bc}} + \frac{b}{{ca}} + \frac{c}{{ab}}} \right) = \sum \frac{(a-b)(a-c)}{a^2} \geqslant 0.$$ Theforere we will show that $$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geqslant 2\sqrt{\left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}} \right)\left( {\frac{a}{c} + \frac{c}{b} + \frac{b}{a}} \right)},$$ equivalent to $$\left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}} \right)+\left( {\frac{a}{c} + \frac{c}{b} + \frac{b}{a}} \right) \geqslant 2\sqrt{\left( {\frac{a}{b} + \frac{b}{c} + \frac{c}{a}} \right)\left( {\frac{a}{c} + \frac{c}{b} + \frac{b}{a}} \right)}.$$ It's $x+y \geqslant 2\sqrt{xy}.$ The proof is completed.

Note. The inequality follows from the famous inequality $$(x+y+z)^2 \geqslant 4(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C).$$

Answer 5

Need to prove$:$ $$\displaystyle \left( {\dfrac {a+b}{c}}+{\dfrac {b+c}{a}}+{\dfrac {a+c}{b}} \right) ^{2 }\geqslant 4 \left( a+b+c \right) \left( {\dfrac {a}{bc}}+{\dfrac {b}{ac}}+{ \dfrac {c}{ab}} \right) $$

Or $$\displaystyle \,{\frac { 2\left( ab-2\,ac+bc \right) ^{2}}{ \left( {a}^{2}+{c}^{2} \right) {b}^{2}}}+{\frac {2 \left( a-c \right) ^{2} \left[2\,b{a}^{2}+abc+2\,b{c}^{2}-ac(a+c) \right]^{2}}{ \left( {a}^{2}+{ c}^{2} \right) \left( {a}^{2}+{b}^{2}+{c}^{2} \right) {c}^{2}{a}^{2}} }+$$ $$+{\dfrac { \left( a-b \right) ^{2} \left( b-c \right) ^{2} \left( a-c \right) ^{2} \left( {a}^{2}+4\,ab+4\,ac+{b}^{2}+4\,bc+{c}^{2} \right) }{ \left( {a}^{2}+{b}^{2}+{c}^{2} \right) {c}^{2}{a}^{2}{b}^{ 2}}}\geqslant 0$$

Which is true.

Text for the SOS's expression: Click here.