prove that $\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$

Question

prove that

$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$$

where $a,b,c>0$ and $n\ge0,n\le 3$,$a+b+c=ab+bc+ca$

My try: by given condition $a+b+c=ab+bc+ca$

we have $a+b+c\le a^2+b^2+c^2$

also using titu's lemma

$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge a+b+c +\frac{3n}{a^2+b^2+c^2}$$ or $$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge ab+bc+ca +\frac{3n}{a^2+b^2+c^2}$$

i dont know what to do next. Any ideas preferably using am-gm.

source Samin Riasat Basics in Olympiad ineq.

Answer 1

We can use also, the following much more weaker estimation: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{(a^2+b^2+c^2)(a+b+c)}{ab+ac+bc},$$ which is just $$\sum_{cyc}(a^2c-b^2a)^2\geq0.$$

Thus, it's enough to prove that: $$\frac{(a^2+b^2+c^2)(a+b+c)}{ab+ac+bc}+\frac{9(ab+ac+bc)^3}{(a^2+b^2+c^2)(a+b+c)^3}\geq\frac{6(ab+ac+bc)}{a+b+c}.$$ Now, let $a^2+b^2+c^2=k(ab+ac+bc)$ again.

Thus, we need to prove that: $$k+\frac{9}{k(k+2)^2}\geq\frac{6}{k+2},$$ which is true by AM-GM: $$k+\frac{9}{k(k+2)^2}\geq2\sqrt{k\cdot\frac{9}{k(k+2)^2}}=\frac{6}{k+2}.$$

Answer 2

We can use the following estimation. $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}.$$ The proof you can see here: https://artofproblemsolving.com/community/c6h296853

For hardest case $n=3$ we need to prove that: $$\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}+\frac{9}{a^2+b^2+c^2}\geq6$$ or $$\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}+\frac{9(ab+ac+bc)^3}{(a^2+b^2+c^2)(a+b+c)^3}\geq\frac{6(ab+ac+bc)}{a+b+c}.$$ Now, let $a^2+b^2+c^2=k(ab+ac+bc).$

Thus, $k\geq1$ and we need to prove that: $$\frac{37k-19}{6}+\frac{9}{k(k+2)}\geq6$$ or $$37k^4+74k^3-55k^2-110k+54\geq0$$ or $$37k^4-37k^3+111k^3-111k^2+56k^2-56k-54k+54\geq0$$ or $$(k-1)(37k^3+111k^2+56k-54)\geq0,$$ which is obvious.

Answer 3

We only need to prove this inequality for $n=3.$ Write the inequality as homogeneous form$:$ $${\dfrac {{a}^{2}}{b}}+{\dfrac {{b}^{2}}{c}}+{\dfrac {{c}^{2}}{a}}+{ \dfrac { 9\left( ab+bc+ac \right) ^{3}}{ \left( {a}^{2}+{b}^{2}+{c}^{2} \right) \left( a+b+c \right) ^{3}}}\geqslant {\dfrac {6(ab+bc+ac)}{a+b+c}}\quad (\text{1})$$

Since the inequality $(\text{1})$ is homogeneous$,$ we may assume $a+b+c=1.$

Let $ab+bc+ca=\frac{1-t^2}{3} \quad (t\in [\,0,1\,]),abc=r.$

We can prove$:$ $${\dfrac {{a}^{2}}{b}}+{\dfrac {{b}^{2}}{c}}+{\dfrac {{c}^{2}}{a}}\geqslant {\dfrac {3\,{t}^{4}-4\,{t}^{3}-5\,{t}^{2}-2\,t-1}{ \left( t-1 \right) \left( 1+2\,t \right) \left( t+1 \right) }} (\text{2}),$$

You can prove this inequality by yourself, I will post my proof for it in another question.

Therefore we need to prove$:$ $${\frac {3\,{t}^{4}-4\,{t}^{3}-5\,{t}^{2}-2\,t-1}{ \left( t-1 \right) \left( 1+2\,t \right) \left( t+1 \right) }}+2\,({t}^{2}-1)-{ \frac { \left( t-1 \right) ^{3} \left( t+1 \right) ^{3}}{1+2\,{t}^{2}} }\geqslant 0.$$

I think now, you can prove it very easy! And the inequality $(\text{2})$ is strong!