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If $a,b,c\ge 0: a+b+c=3$ then prove$$\frac{a}{(b^2+3)^2}+\frac{b}{(c^2+3)^2}+\frac{c}{(a^2+3)^2}\ge\frac{3}{16}.$$

I've tried to use Holder inequality as$$\sum_{cyc}a\cdot\sum_{cyc}\frac{a}{(b^2+3)^2}\ge\left(\sum_{cyc}\frac{a}{b^2+3}\right)^2.$$Thus, it remains to show$$\frac{a}{b^2+3}+\frac{b}{c^2+3}+\frac{c}{a^2+3}\ge \frac{3}{4}.$$ By AM-GM $$\frac{3a}{b^2+3}=a-\frac{ab^2}{b^2+3}\ge a-\frac{ab^2}{4\sqrt[4]{b^2}}= a-\sqrt{a^2b^3}. $$ Now, we need to prove$$\sqrt{a^2b^3}+\sqrt{b^2c^3}+\sqrt{c^2a^3}\le 3.$$I don't know how to prove the last inequality.

Hope you can give some hints to prove my rest part also for the original inequality.

Anonymous
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2 Answers2

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Proof.

Notice that$$\frac{9a}{(b^2+3)^2}=a-\frac{ab^4+6b^2a}{(b^2+3)^2},$$we'll prove$$\frac{ab^4+6b^2a}{(b^2+3)^2}+\frac{bc^4+6c^2b}{(c^2+3)^2}+\frac{ca^4+6a^2c}{(a^2+3)^2}\le \frac{21}{16}.$$ By AM-GM and Cauchy-Schwarz inequality$$\frac{ab^4+6b^2a}{(b^2+3)^2}\le \frac{ab^2(b^2+1)+5ab^2}{8(b^2+1)}\le \frac{ab^2}{8}+\frac{5ab}{16}.$$It implies$$\sum_{cyc}\frac{ab^4+6b^2a}{(b^2+3)^2}\le \frac{ab^2+bc^2+ca^2}{8}+\frac{5(ab+bc+ca)}{16}.\tag{1}$$ Also \begin{align*} 3(a^2+b^2+c^2)&=(a+b+c)(a^2+b^2+c^2)\\&=ab^2+bc^2+ca^2+a^3+b^3+c^3+cb^2+ac^2+ba^2\\&=b(b-a)^2+c(c-b)^2+a(a-c)^2+3(ab^2+bc^2+ca^2)\\&\ge 3(ab^2+bc^2+ca^2). \end{align*} Thus, $$a^2+b^2+c^2\ge ab^2+bc^2+ca^2.\tag{2}$$ From $(1),$ $(2)$ and $3\ge ab+bc+ca$ we obtain $$\sum_{cyc}\frac{ab^4+6b^2a}{(b^2+3)^2}\le \frac{2(a^2+b^2+c^2)+5(ab+bc+ca)}{16}=\frac{18+ab+bc+ca}{16}\le \frac{21}{16}.$$ Hence, the proof is done. Equality holds iff $a=b=c=1.$

Remark. I check your last inequality and it's true but I haven't seen a nice proof yet.

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Alternative proof.

Thank to your attempt, we'll prove $$\sqrt{a^2b^3}+\sqrt{b^2c^3}+\sqrt{c^2a^3}\le 3.$$ By using CBS inequality as $$\sum_{cyc}\sqrt{a^2b^3}=\sum_{cyc}\sqrt{ab\cdot ab^2}\le \sqrt{(ab+bc+ca)(ab^2+bc^2+ca^2)}.$$ Thus, it's enough to prove $$(ab+bc+ca)(ab^2+bc^2+ca^2)\le 9. \tag{*}$$ It was discussed here

We may use BW here.

After homogenization, $(*)$ becomes $$(a+b+c)^5\geq 27(ab+bc+ca)(ab^2+bc^2+ca^2).$$

Equality holds iff $a=b=c.$

WLOG, assuming that $a=\min\{a,b,c\}.$ Let $b=a+u, c=a+v$ where $u,v\ge 0.$

Now, we obtain \begin{align*} (a+b+c)^5-27(ab+bc+ca)(ab^2+bc^2+ca^2) \end{align*} $$=27a^3(u^2-uv+v^2)+9a^2(4u^3-6u^2v+3uv^2+4v^3)+3a(5u^4-7u^3v-6u^2v^2+11uv^3+5v^4)+u^5+5u^4v+10u^2v^3+5uv^4+v^5\ge 0.$$

Hence, the proof of $(*)$ is done.

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