Prove that $\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$

Question

If $a+b+c=1$ and $a,b,c>0$ then prove that $$\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$$

My try : We have to prove: $$ab+bc+ca+36abc(ab+bc+ca)\ge 21abc$$ or after homogenising we get : $$\sum a^4b+3\sum a^3b^2+6\sum a^2b^2c\ge 14\sum a^3bc$$ I don't know what to do next. I think SOS may be helpful but find it difficult to factorize. Any other methods are also welcome but, if possible, can someone help me continue from here?

Answer 1

Now, we need to prove that: $$\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2)+4\sum_{cyc}(a^3b^2+a^3c^2-2a^3bc)-6\sum_{cyc}(a^3bc-a^2b^2c)\geq0$$ or

$$\sum_{cyc}(a^4b-a^3b^2-a^2b^3+ab^4)+$$ $$+4\sum_{cyc}(c^3a^2+c^3b^2-2c^3bc)-3abc\sum_{cyc}(a^2+b^2-2ab)\geq0$$ or $$\sum_{cyc}(a-b)^2(ab(a+b)+4c^3-3abc)\geq0,$$ which is true by AM-GM: $$ab(a+b)+4c^3\geq2\sqrt{a^3b^3}+c^3\geq3\sqrt[3]{a^3b^3c^3}=3abc$$ The following inequality a bit of stronger.

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=1$. Prove that: $$\frac{1}{abc}+48\geq\frac{25}{ab+ac+bc}.$$

Answer 2

Another way.

Let $a=\min\{a,b,c\},$ $b=a+u$ and $c=a+v$.

Thus, $u\geq0$, $v\geq0$ and $$\sum_{cyc}(a^4b+a^4c+3a^3b^2+3a^3c^2-14a^3bc+6a^2b^2c^2)=$$ $$=6(u^2-uv+v^2)a^3+9uv(u+v)a^2+$$ $$+2(u^4-2u^3v+12u^2v^2-2uv^3+v^4)a+uv(u+v)^3\geq0.$$

Answer 3

I think, the shortest way here it's using $uvw$.

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $f(w^3)\geq0,$ where $$f(w^3)=\frac{3u^3}{w^3}+4-\frac{7u^2}{v^2}.$$ Now we see that $f$ decreases, which says that it's enough to prove our inequality for a maximal value of $w^3$, which by $uvw$ happens for equality case of two variables.

We could have seen it before because the starting inequality it's a polynomial symmetric inequality of fifth degree.

Now, after homogenization we need to prove that: $$\frac{(a+b+c)^3}{abc}+36\geq\frac{21(a+b+c)^2}{ab+ac+bc}.$$ Since the last inequality is homogeneous and symmetric, it's enough to assume $b=c=1$, which gives: $$(a-1)^2(a^2-2a+4)\geq0,$$ which is obvious.

Answer 4

Suppose $t = \frac{a+b}{2}$ and $c = \max \{a,b,c\}.$ Let $$f(a,b,c) = \frac{1}{abc}+36 - \frac{21}{ab+bc+ca}.$$ We have $$f(a,b,c) -f(t,t,c) = \frac{1}{abc}-\frac{4}{c(a+b)^2}+\frac{84}{(a+b)(a+b+4c)}-\frac{21}{ab+bc+ca}$$ $$ = \frac{(a-b)^2}{a+b}\left(\frac{1}{abc(a+b)}-\frac{21}{(ab+bc+ca)(a+b+4c)}\right) \geqslant 0,$$ because $$(ab+bc+ca)(a+b+4c)=(a+b+c)(ab+bc+ca)(a+b+4c) $$ $$ \geqslant 9abc (a+b+4c) > 21abc(a+b).$$ Thefore $$f(a,b,c) \geqslant f(t,t,c) = f\left(\frac{1-c}{2},\frac{1-c}{2},c\right)=\frac{4(3c^2-3c+1)(3c-1)^2}{c(3c+1)(c-1)^2} \geqslant 0.$$ The proof is completed.

Answer 5

I found another SOS proof$:$

After homogenising the inequality become$:$ $$a \left( b-c \right) ^{4}+b \left( c-a\right) ^{4}+c \left( a-b \right) ^{4}+3\,{a}^{3} \left( b-c \right) ^{2}+3\,{b}^{3} \left( c-a \right) ^{2}+3\,{c}^{3} \left( a-b \right) ^{2}\geqslant 0$$

Or another SOS$:$ $$\sum \left( {a}^{2}b+a{b}^{2}-6\,abc+3\,{c}^{2}a+3\,{c}^{2}b+{c}^{3} \right) \left( a-b \right) ^{2}\geqslant 0$$

which is easy to prove$:$ $$ {a}^{2}b+a{b}^{2}-6\,abc+3\,{c}^{2}a+3\,{c}^{2}b+{c}^{3} \geqslant 0.$$ Can you end it now?