If $a,b,c >0 : ab+bc+ca=3,$ find maximal value $\sum\dfrac{a\sqrt{a^2+2}}{a^2+3}$

Question

Question

If $a,b,c >0 : ab+bc+ca=3,$ find maximal value $$M=\dfrac{a\sqrt{a^2+2}}{a^2+3}+\dfrac{b\sqrt{b^2+2}}{b^2+3}+\dfrac{c\sqrt{c^2+2}}{c^2+3}$$ By $a=b=c=1,$ I try prove $M\le \dfrac{3\sqrt{3}}{4}$ or $$\sum_{cyc}(ab+ac)\sqrt{a^2+2}\le \dfrac{3\sqrt{3}}{4}(a+b)(b+c)(c+a)$$ I don't know how to find upbound $\sum_{cyc}(ab+ac)\sqrt{a^2+2}$ I need some advices. Thanks.

Answer 1

Let $a=\sqrt3\tan\frac{\alpha}{2},$ $b=\sqrt3\tan\frac{\beta}{2}$ and $c=\sqrt3\tan\frac{\gamma}{2},$ where $\{\alpha,\beta,\gamma\}\subset(0,\pi).$

Thus, $\sum\limits_{cyc}\tan\frac{\alpha}{2}\tan\frac{\beta}{2}=1,$ which gives $\alpha+\beta+\gamma=\pi$ and we need to prove that: $$\sum_{cyc}\frac{\sqrt3\tan\frac{\alpha}{2}\sqrt{3\tan^2\frac{\alpha}{2}+2}}{3\tan^2\frac{\alpha}{2}+3}\leq\frac{3\sqrt3}{4}$$ or $$\sum_{cyc}\sin\frac{\alpha}{2}\sqrt{3\sin^2\frac{\alpha}{2}+2\cos^2\frac{\alpha}{2}}\leq\frac{9}{4}$$ or $$\sum_{cyc}\sin\frac{\alpha}{2}\sqrt{\sin^2\frac{\alpha}{2}+2}\leq\frac{9}{4}.$$

Now, let $f(x)=\sin\frac{x}{2}\sqrt{\sin^2\frac{x}{2}+2}.$ But $$f''(x)=\frac{2\sin\frac{x}{2}\left(\cos^2x-7\cos{x}+4\right)}{\sqrt{\left(\sin^2\frac{x}{2}+2\right)^3}},$$ which says that $\arccos\frac{7-\sqrt{33}}{2}$ is an unique inflection point of $f$ on $(0,\pi)$, which by the Vasc's HCF Theorem it's enough to prove our inequality for equality case of two variables.

Now, the homogenization gives $$\sum_{cyc}a(b+c)\sqrt{3a^2+2(ab+ac+bc)}\leq\frac{9}{4}\prod_{cyc}(a+b),$$ which is obvious for $b=c=0$.

Bu for $b=c=1$ we need to prove that: $$2a\sqrt{3a^2+4a+2}+2(a+1)\sqrt{4a+5}\leq\frac{9}{2}(a+1)^2$$ and the rest is smooth.

Can you end it now?

About HCF see here: https://www.scribd.com/document/399137224/Vasile-Cirtoaje-Mathematical-Inequalities-Vol-4#

I got that finally we need to prove that: $$(a-1)^2(1089a^6+2826a^5+2895a^4+1548a^3+655a^2+202a+1)\geq0$$ and we see that $(1,1,0)$ makes a trouble.

Answer 2

Another way.

After homogenization we need to prove that: $$\sum_{cyc}a(b+c)\sqrt{3a^2+2(ab+ac+bc)}\leq\frac{9}{4}\prod_{cyc}(a+b).$$ But by C-S $$\sum_{cyc}a(b+c)\sqrt{3a^2+2(ab+ac+bc)}\leq\sqrt{(a+b+c)\sum_{cyc}a(b+c)^2(3a^2+2(ab+ac+bc)}$$ and it's enough to prove that: $$\frac{81}{16}\prod_{cyc}(a+b)^2\geq(a+b+c)\sum_{cyc}a(b+c)^2(3a^2+2(ab+ac+bc)$$ or $$\left(\sum_{cyc}(a^2b+a^2c-2abc)\right)^2\geq0.$$

Answer 3

Proof.

By AM-GM,$$\dfrac{(a^2+2)(b+c)}{(a+b)(c+a)}+\dfrac{12}{(a+b)(b+c)(c+a)} \ge \dfrac{4\sqrt{3}\cdot\sqrt{a^2+2}}{(a+b)(c+a)},$$ $$\implies\dfrac{4\sqrt{3}\cdot a\sqrt{a^2+2}}{a^2+3} \le \dfrac{a(12+(a^2+2)(b+c)^2)}{(a+b)(b+c)(c+a)}.$$ Similarly, we will prove $$\sum\limits_{cyc}a(12+(a^2+2)(b+c)^2) \le 9(a+b)(b+c)(c+a).$$ Let $p=a+b+c, q=ab+bc+ca=3, r=abc$, we obtain \begin{align*} \sum\limits_{cyc}a(12+(a^2+2)(b+c)^2)&=pq^2+2pq-5qr+12p+6r\\ &=27p-9r\\ &=9(a+b)(b+c)(c+a). \end{align*} The proof is done. Equality holds at $a=b=c=1.$

Answer 4

Alternative proof.

It is easy to rewrite the OP as $$\sum_{cyc}\frac{a}{(a+b)(a+c)}\sqrt{3a^2+2(ab+bc+ca)}\le \frac{9}{4},$$or$$\sum_{cyc}\frac{2a}{\sqrt{(a+b)(a+c)}}.\sqrt{8+\frac{4a^2}{(a+b)(a+c)}}\le 9.$$ Now, denoting $$x=\frac{2a}{\sqrt{(a+b)(a+c)}};y=\frac{2b}{\sqrt{(a+b)(b+c)}};z=\frac{2c}{\sqrt{(c+b)(a+c)}}.$$It implies that $x^2+y^2+z^2+xyz=4$ and the remain is proving $$x\sqrt{x^2+8}+y\sqrt{y^2+8}+z\sqrt{z^2+8}\le 9. \tag{*}$$Apply AM-GM as $$x.\sqrt{x^2+8}\le \frac{x}{2}.\left(\frac{x^2+8}{2x+yz}+2x+yz\right).$$Thus, we obtain $$2.LHS_{(*)}\le \sum_{cyc}\frac{x^3+8x}{2x+yz}+xyz+8.$$We can used a well-known identity $\dfrac{x}{2x+yz}+\dfrac{y}{2y+xz}+\dfrac{z}{2z+yx}=1,$ which easily to verify according to $x^2+y^2+z^2+xyz=4.$

Indeed, we will prove $$\sum_{cyc}\frac{x^3+8x}{2x+yz}+xyz+8=18,$$or $$\frac{x^3}{2x+yz}+\frac{y^3}{2y+zx}+\frac{z^3}{2z+xy}=2-xyz,$$ which is true since$$\sum_{cyc}\dfrac{2x^3}{2x+yz}=\sum_{cyc}\left(x^2-\dfrac{x^2yz}{2x+yz}\right)=x^2+y^2+z^2-xyz.\sum_{cyc}\frac{x}{2x+yz}=4-2xyz.$$Hence, the $(*)$ is proven and we end proof here.