# Proving $\frac{1}{abc}+\frac{12}{a^2b+b^2c+c^2a}\ge5$

Let $$a,b,c>0$$, $$a+b+c=3$$. Prove that$$\frac1{abc}+\frac{12}{a^2b+b^2c+c^2a}\ge5$$

My approach using a well-known result:$$a^2b+b^2c+c^2a+abc\le\frac4{27}(a+b+c)^3$$ We need to prove that $$\frac1{abc}+\frac{12}{4-abc}\ge5$$ but this inequality does not hold for all $$a,b,c$$.

Is there any better idea to help me solve the problem? Thanks for your help!

• Is there better lemma for $a^2b+b^2c+c^2a$?
– Mars
Dec 25, 2021 at 1:45
• Any one help me?
– Mars
Dec 26, 2021 at 10:48

Remark $$\displaystyle\sum f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$$ means cyclic sum.

Homogenize it, then we get

denote \begin{aligned} f\left( a,b,c \right) :=&\sum{a^5b}+3\sum{a^4b^2}+3\sum{a^3b^3}+\sum{a^2b^4}\\ &+15\sum{a^4bc}-92\sum{a^3b^2c}+42\sum{a^2b^3c}+81a^2b^2c^2 \end{aligned}

we gonna prove $$f(a,b,c)\geq 0$$

actually, we have \begin{aligned} &\left( a+b+c \right) ^2f\left( a,b,c \right)\\ =& 28\sum{a^2b^2c}\sum{a\left( a-b \right) \left( a-c \right)}+\frac{13}{24}\sum{\left( a^2b-abc \right)^2}\sum{\left( a-b \right) ^2}\\ &+\frac{59}{12}\sum{\left( a^2b-abc \right)}\sum{\left( a-b \right) ^2b^2c}+\frac{65}{4}\sum{a^2bc\left( -a^2+ba+ca+b^2-2bc \right) ^2}\\ &+7\sum{a^3b^2\left( b-c \right) ^2c}+\frac{15}{2}\sum{a^3b\left( b^2-ac \right) ^2}+\frac{83}{12}\sum{a^4b^2\left( b-c \right) ^2}\\ &+\frac{121}{12}\sum{a\left( a-b \right) ^2\left( b-c \right) ^2c^3}+\frac{17}{4}\sum{ab^2c\left( -a^2+ba+ca+b^2-2bc \right) ^2}\\ &+\frac{1}{2}\sum{\left( a-b \right) ^4\left( ba^3-3b^2ca+b^2c^2+b^3c \right)}\\ &+\frac{253}{12}\sum{abc^2\left( -a^2+ba+ca+b^2-2bc \right) ^2}+\frac{1}{2}\sum{\left( a^2-2ba+b^2-bc \right) ^2\left( c^2-ab \right) ^2}\\ \geq & 0 \end{aligned}

• Very good! Thanks
– Mars
Mar 29, 2022 at 7:59

it clear this inequality is equivalent to: prove $$a+b+c=1$$,then $$\dfrac{1}{abc}+\dfrac{12}{a^2b+b^2c+c^2a}\ge 135$$Note $$2(a^2b+b^2c+c^2a)=\sum ab(a+b)+(a-b)(b-c)(a-c)\le \sum ab(a+b)+\sqrt{(a-b)^2(b-c)^2(c-a)^2}$$ and $$\sum ab(a+b)=(a+b+c)(ab+bc+ac)-3abc=q-3r$$

$$(a-b)^2(b-c)^2(c-a)^2=q^2-4q^3+2(9q-2)r-27r^2$$

where $$p=a+b+c=1,q=ab+bc+ca,r=abc$$ then we have $$\frac{1}{r}+\frac{24}{q-3r+\sqrt{q^2-4q^3+2(9q-2)r-27r^2}} \ge 135 .$$ if$$r\le\dfrac{1}{135}$$ then this is obvious true

if $$\frac{1}{135} ,this inequality is equivalent to: $$\left(\frac{405r^2+21r}{135r-1}-q \right)^2-(q^2-4q^3+2(9q-2)r-27r^2) \ge 0.$$ or $$4q^3-\frac{24r(135r+1)}{135r-1} q +4r+27r^2+\left(\frac{405r^2+21r}{135r-1}\right)^2 \ge 0.$$ since use AM-GM $$4q^3+8 \sqrt{\left(\frac{2r(135r+1)}{135r-1}\right)^3} \ge \frac{24r(135r+1)}{135r-1} q.$$ It suffices to show that: $$4+27r+\frac{r(21+405r)^2}{(135r-1)^2} \ge 8 \sqrt{\frac{8r(135r+1)^3}{(135r-1)^3}}.$$ i.e $$1+\frac{9r((135r+1)^2+12)}{(135r-1)^2} \ge 2\sqrt{\frac{8r(135r+1)^3}{(135r-1)^3}}.$$ let $$t=\dfrac{135r+1}{135r-1}=1+\dfrac{2}{135r-1}\ge\dfrac{3}{2}$$,

then The last inequality becomes: $$1+\frac{9(t+1)(4t^2-6t+3)}{135(t-1)} \ge 2\sqrt{\frac{8t^3(t+1)}{135(t-1)}} .$$ $$\Longleftrightarrow 81(2t^3-t^2+6t-6)^2\ge 1080t^3(t^2-1)$$ or $$27(2t-3)^2(3t^4-4t^3-8t+12)\ge 0,t\ge\frac{3}{2}$$ it is clear, because $$3t^4-4t^3-8t+12=\dfrac{81}{16}(2t-3)^4+\dfrac{189}{4}(2t-3)^3+\dfrac{1215}{8}(2t-3)^2+\dfrac{297}{4}(2t-3)+\dfrac{729}{16}\ge 0$$

• It is impposible for me to think of it! Thanks
– Mars
Mar 29, 2022 at 8:01