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I'm looking for some ideas to deal with my inequality.

P/s: I'd like to create simple look inequalities so we can call them as artificial inequalities. I agree that my inequality may be wrong but I checked it carefully before posting here.

Problem. Let $a,b,c\ge 0: ab+bc+ca>0$ and $a+b+c=2.$ Prove that$$\color{black}{\frac{a\sqrt{b+bc+c}}{b+c}+\frac{b\sqrt{c+ca+a}}{c+a}+\frac{c\sqrt{a+ab+b}}{a+b}\ge 2.}$$

Source: own

The equality occurs for $(a,b,c)\sim\left(\dfrac{2}{3},\dfrac{2}{3},\dfrac{2}{3}\right)$ or $(a,b,c)\sim\left(1,1,0\right).$

I found that the following Holder using helps.

Indeed, by Holder inequality$$\left(\sum_{\mathrm{cyc}}\frac{a\sqrt{b+bc+c}}{b+c}\right)^2\cdot\sum_{\mathrm{cyc}}\frac{a(b+c)^2(a+1)^3}{b+bc+c}\ge \left[\sum_{\mathrm{cyc}}a(a+1)\right]^3.$$It's enough to prove$$\left(a^2+b^2+c^2+2\right)^3\ge 4\sum_{\mathrm{cyc}}\frac{a(b+c)^2(a+1)^3}{b+bc+c}$$which is true but my proof for the last inequality was not nice.

Hope someone share other approaches. Thank you for your interest.

Tran Ngoc Khuong Trang
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1 Answers1

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We need to prove $$\sum_{cyc}\frac{a\sqrt{b^2+c^2+ab+ac+4bc}}{b+c}\geq\sqrt2(a+b+c)$$ or $$\sum_{cyc}\left(\tfrac{a^2(b^2+c^2+ab+ac+4bc)}{(b+c)^2}+\tfrac{2ab\sqrt{(a^2+c^2+ab+bc+4ac)(b^2+c^2+ab+ac+4bc)}}{(a+c)(b+c)}\right)\geq2(a+b+c)^2.$$ Now, by C-S and AM-GM $$\sqrt{(a^2+c^2+ab+bc+4ac)(b^2+c^2+ab+ac+4bc)}\geq ab+c^2+ab+ac+bc+3c\sqrt{ab}\geq$$ $$\geq2ab+c^2+ac+bc+\frac{6abc}{a+b}$$ and it's enough to prove that: $$\sum_{cyc}\left(\tfrac{a^2(b^2+c^2+ab+ac+4bc)}{(b+c)^2}+\tfrac{2ab\left(2ab+c^2+ac+bc+\frac{6abc}{a+b}\right)}{(a+c)(b+c)}\right)\geq2(a+b+c)^2$$ or $$\sum_{cyc}\left(\tfrac{a^2(ab+ac+2bc)}{(b+c)^2}+\tfrac{4a^2b^2}{(a+c)(b+c)}+\tfrac{2abc((a+b)(a+b+c)+6ab}{(a+b)(a+c)(b+c)}\right)\geq\sum_{cyc}(a^2+4ab),$$ which is true by SOS.

Indeed, after full expanding we need to prove that: $$\sum_{sym}(a^7b+a^6b^2-a^5b^3c-a^4b^4+2a^6bc-a^5b^3c-4a^4b^3c+3a^3b^3c^2)\geq0.$$ Now, $a\geq b\geq c$.

Thus, $$\sum_{sym}(a^7b+a^6b^2-a^5b^3-a^4b^4+2a^6bc-a^5b^3c-4a^4b^3c+3a^3b^3c^2)=$$ $$=\sum_{sym}(a^7b-a^6b^2+2a^6b^2-2a^5b^3+a^5b^3-a^4b^4)+$$ $$+abc\sum_{sym}(2a^5-2a^4b+a^4b-a^3b^2-3a^3b^2+3a^2b^2c)=$$ $$=\sum_{cyc}(a-b)^2ab(a^4+a^3b+a^2b^2+ab^3+b^4+2a^3b+2a^2b^2+2ab^3+2c(a+b)(a^2+b^2)+abc(a+b)-3c^3(a+b))\geq$$ $$\geq\sum_{cyc}(a-b)^2(a+b)ab((a+b)(a^2+b^2)+c(2a^2+2b^2+ab)-3c^3)\geq$$ $$\geq\sum_{cyc}(a-b)^2(a+b)ab(a^3+b^3-c^3+2c(a^2+b^2-c^2))\geq$$ $$\geq(a-c)^2(a+c)ac(a^3+c^3-b^3)+(b-c)^2(b+c)bc(b^3+c^3-a^3)+$$ $$+2abc((a-c)^2(a+c)(a^2+c^2-b^2)+(b-c)^2(b+c)(b^2+c^2-a^2))\geq$$ $$\geq(b-c)^2(a+c)ac(a^3-b^3)+(b-c)^2(b+c)bc(b^3-a^3)+$$ $$+2abc((b-c)^2(a+c)(a^2-b^2)+(b-c)^2(b+c)(b^2-a^2))\geq0.$$

Michael Rozenberg
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