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