What is min $T=\sqrt{\frac{1-bc}{2b^2+5bc+2c^2}}+\sqrt{\frac{1-ac}{2a^2+5ac+2c^2}}+\sqrt{\frac{1-ba}{2b^2+5ba+2a^2}}$

Question

Find the minimal value $$P=\sqrt{\frac{1-bc}{2b^2+5bc+2c^2}}+\sqrt{\frac{1-ac}{2a^2+5ac+2c^2}}+\sqrt{\frac{1-ba}{2b^2+5ba+2a^2}}$$ when $a,b,c\ge 0: ab+bc+ca=1.$

I set $a=b=1;c=0$ I got $P=\sqrt{2}.$

By AM-GM $$P=\sum_{cyc}\frac{1-bc}{\sqrt{(1-bc)(2b^2+2c^2+5bc)}}\ge \sum_{cyc}\frac{2(1-bc)}{(1-bc)+(2b^2+2c^2+5bc)}=\sum_{cyc}\frac{1-bc}{(b+c)^2}\ge\sqrt{2} $$ I stopped here.I'll be appreciate if someone give me a good approach!

Answer 1

Proof.

We'll prove the minimum is $\sqrt{2}$ and attained at $(0,1,1).$

It means that $$\sqrt{\frac{a(b+c)}{2b^2+5bc+2c^2}}+\sqrt{\frac{b(a+c)}{2a^2+5ac+2c^2}}+\sqrt{\frac{c(a+b)}{2b^2+5ba+2a^2}}\ge \sqrt{2}.$$ By using Holder $$\left(\sum_{cyc}\sqrt{\frac{a(b+c)}{2b^2+5bc+2c^2}}\right)^2.\sum_{cyc}\frac{a^2}{b+c}(2b^2+5bc+2c^2)\ge (a+b+c)^3.$$Thus, it remains to prove that$$(a+b+c)^3\ge 2\sum_{cyc}\frac{a^2}{b+c}(2b^2+5bc+2c^2), $$or $$\sum_{cyc}a^3-\sum_{cyc}a^2(b+c)-2abc\sum_{cyc}\frac{a}{b+c}\ge 0,$$or $$\sum_{cyc}a^3+3abc-\sum_{cyc}a^2(b+c)-2abc\sum_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)\ge 0. \tag{*}$$ Notice that we easily get identities $$a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\sum_{cyc}b-c)^2;$$$$6abc-\sum_{cyc}a^2(b+c)=-a(b-c)^2-b(c-a)^2-c(a-b)^2;$$ \begin{align*} \sum_{cyc}\frac{a}{b+c}-\frac{3}{2}&=\frac{1}{2(a+b)(b+c)(c+a)}.\sum_{cyc}a^2(2a-b-c)\\&=\frac{1}{2}.\sum_{cyc}\frac{(b-c)^2}{(a+b)(a+c)}. \end{align*} Hence, the $(*)$ is equivalent to$$\sum_{cyc}(b-c)^2\left(b+c-a-\frac{2abc}{(a+b)(a+c)}\right)\ge 0,$$which $SOS$ kills it.

About $SOS$ see here.

Can you go further ?

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Indeed, denoting $$S_a=b+c-a-\frac{2abc}{(a+b)(a+c)};$$$$S_b=a+c-b-\frac{2abc}{(a+b)(b+c)};$$$$S_c=b+a-c-\frac{2abc}{(c+b)(a+c)};$$and the $(**)$ turns out $$S_a(b-c)^2+S_b(c-a)^2+S_c(a-b)^2\ge 0.$$WLOG, assuming that $a\ge b\ge c\ge 0.$ Notice that $$S_b=\frac{b(a^2+c^2-b^2)+ac(a+c-b)}{(a+b)(b+c)(c+a)}\ge 0.$$Thus, by $SOS$ theorem we need to prove $S_a+S_b\ge 0; S_c+S_b\ge 0.$

• $S_a+S_b\ge 0.$ It means that $$2c\left[1-\frac{ab}{(a+b)}.\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\right]=2c\left(1-\frac{ab(2c+a+b)}{(a+b)(b+c)(c+a)}\right)\ge 0.$$
• $S_c+S_b\ge 0.$ It means that $$2a\left[1-\frac{bc}{(c+b)}.\left(\frac{1}{a+c}+\frac{1}{b+a}\right)\right]=2a\left(1-\frac{bc(2a+c+b)}{(a+b)(b+c)(c+a)}\right)\ge 0.$$ Hence $(**)$ is proven and the proof is done.

Answer 2

By Holder $$\left(\sum_{cyc}\sqrt{\frac{a(b+c)}{2b^2+5bc+2c^2}}\right)^2\sum_{cyc}\frac{a^2(2b^2+5bc+2c^2)}{b+c}\geq(a+b+c)^3.$$ Thus, it's enough to prove that $$(a+b+c)^3\geq2\sum_{cyc}\frac{a^2(2b^2+5bc+2c^2)}{b+c}$$ and the rest is smooth:

We need to prove that $$\sum_{sym}(a^5b-a^3b^3-a^4bc+a^3b^2c)\geq0.$$ Now, let $a\geq b\geq c.$

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