Equivalent to show $(\sum ab)(\sum a)\geq 8+\prod a$ ,Now $(\sum ab)\geq 3$ so now it's enough to show $3x^2\geq 8x+3$, true surely, where $x=\sum a$ ,So $\sum a=3\implies \sum ab=3 \implies a=b=c=1$ is the equality case.

Note that $(ab+bc+ca)^2\geq 3abc(a+b+c)=9,$ so $ab+bc+ca\geq 3.$ Therefore, by the AM-GM inequality we have \[(a+b)(b+c)(c+a)\geq \frac 89(a+b+c)(ab+bc+ca)\geq \frac 89\cdot 3\cdot 3 = 8;\] Since $a+b+c\geq \sqrt{3(ab+bc+ca)}\geq 3.$ Hence we are done. Equality holds iff $a=b=c=1.\Box$

$3 = abc(a+b+c) \ge abc \cdot 3 \sqrt[3]{abc} = 3(abc)^{\frac43} \Rightarrow abc \le 1$ $\Rightarrow -abc \ge 1$ and $\sqrt[3]{\frac1{abc}} \ge 1$ Now $(a+b)(b+c)(c+a) = (a+b+c)(ab+bc+ca)-abc = \frac3{abc} \cdot (ab+bc+ca) - abc = 3\left(\frac1a+\frac1b+\frac1c\right) - abc \ge 3 \cdot 3 \sqrt[3]{\frac1{abc}} -abc \ge 9(1) - 1 = 8$. Equality iff all inequalities above are equalities iff $a=b=c=1$.

There is another good solution. Substitute a+b=x and so on. we have to use herons formula and area =abc/4R It turns into a geometric inequality which is solvable

Shravu wrote: There is another good solution. Substitute a+b=x and so on. we have to use herons formula and area =abc/4R It turns into a geometric inequality which is solvable Shravan bal, asa naye karu

$(a+b)(b+c)(c+a) = (a+b)(ab+(a+b+c)c)$ $= (a+b)(ab+\frac1{ab}+\frac1{ab}+\frac1{ab})\ge2\sqrt{ab}\cdot 4\sqrt[4]{\frac1{a^2b^2}}=8$.

Let $a,b,c$ be positive real numbers such that $abc(a+b+c)=3.$ Prove that\[(a+b)(b+c)(c+a)\geq 9-abc\ge8.\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=511245&p=2870626#p2870626 Let $a,b,c$ be positive real numbers such that $abc(a+b+c)=3.$ Prove that\[(a^2+bc)(b^2+ca)(c^2+ab)\geq 9-a^2b^2c^2.\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2871454

Let $x=b+c, y=c+a, z=a+b$. Then $x,y,z$ are the lengths of sides of a triangle $\triangle XYZ$. Let $s, \Delta, R$ be the semiperimeter, area, circumradius respectively of $\triangle XYZ$. Then we have $s = \frac12(x+y+z) = a+b+c; \ s-x = a;\ s-y = b;\ s-z=c$. Thus by Heron's formula, $\Delta^2 = s(s-x)(s-y)(s-z) = abc(a+b+c) = 3 \Rightarrow \Delta = \sqrt{3}$. By the well known inequality $\frac{3\sqrt{3}}{4} R^2 \ge \Delta$ we get $\frac{3\sqrt{3}}{4}R^2 \ge \sqrt3 \Rightarrow R \ge \frac{2}{\sqrt3}$. Again we have $\Delta = \dfrac{xyz}{4R} \Rightarrow xyz=4R\Delta$. Thus $(a+b)(b+c)(c+a) = xyz = 4R\Delta \ge 4 \cdot \frac{2}{\sqrt3} \cdot \sqrt3 = 8$, proving the required inequality.

sqing wrote: $(a+b)(b+c)(c+a) = (a+b)(ab+(a+b+c)c)$ $= (a+b)(ab+\frac1{ab}+\frac1{ab}+\frac1{ab})\ge2\sqrt{ab}\cdot 4\sqrt[4]{\frac1{a^2b^2}}=8$. Nice.

this is equivalent to $\sin A\sin B\sin C\le\frac{3\sqrt{3}}{8}$

sankha012 wrote: this is equivalent to $\sin A\sin B\sin C\le\frac{3\sqrt{3}}{8}$ yeah. $sinA=\frac{2\sqrt{xyz(x+y+z)}}{(z+x)(x+y)}.$

cmtappu96 wrote: Shravu wrote: There is another good solution. Substitute a+b=x and so on. we have to use herons formula and area =abc/4R It turns into a geometric inequality which is solvable Shravan bal, asa naye karu cmtappu: fyi, mala to problem parikshet tya method ni ardha alela, ase ka naye karu?

Best way to be good at inequalities is to remember some very useful results.

This is an ugly solution, but whatever. Assume all summations are cyclic. First note that by AM-GM we have $abc(a+b+c)=3\geq abc\cdot 3\sqrt[3]{abc}=3(abc)^{4/3}\implies abc\leq 1$. Hence, we have $9\geq 8(abc)^{1/3}+(abc)^{4/3}$, and multiplying this by $(abc)^{2/3}$ gives $9\sqrt[3]{(abc)^2}\geq 8abc+a^2b^2c^2$. However, note that \[\dfrac13\sum ab\geq\sqrt[3]{a^2b^2c^2}\implies 3\sum ab\geq 9\sqrt[3]{a^2b^2c^2},\] so we have $3\sum ab\geq 8abc+a^2b^2c^2$. Next, use the condition $abc(a+b+c)=3$ to get \begin{align*}\sum ab(a^2bc+ab^2c+abc^2)&=\sum(a^3b^2c+a^2b^3c+a^2b^2c^2)\\&=\sum a^2b^2c(a+b)+3a^2b^2c^2\geq 8abc+a^2b^2c^2.\end{align*} Subtract $a^2b^2c^2$ from both sides. Note that the LHS becomes \begin{align*}\sum(a^3b^2c+a^2b^3c+a^2b^2c^2)&=abc\left[\sum ab(a+b)+2abc\right]\\&=abc(a+b)(ab+ac+bc+c^2)\\&=abc(a+b)(b+c)(c+a).\end{align*} Since the RHS of the inequality is now $8abc$, we have $abc(a+b)(b+c)(c+a)\geq 8abc$. Dividing both sides by $abc$ gives our desired. $\blacksquare$ EDIT: This seems to bear some resemblance to subham's solution, except the fact that I divide out $abc$ toward the end of the solution makes mine much uglier. Also, equality comes trivially when $a=b=c=1$.

please hide solutions!

sankha012 wrote: this is equivalent to $\sin A\sin B\sin C\le\frac{3\sqrt{3}}{8}$ This can be proved by the jensen's inequaltily.(will post full solution later) Aayush

Hey can someone point out the mistake:- $abc(a+b+c)=3.$ Using $A.M.-G.M.$ we have $abc\leq1$ $(a+b)(b+c)(c+a)\geq 8.$ on expanding becomes $(2,1,0)+2abc\geq8$. As $(2,1,0)\succ(1,1,1)$ we get $8abc\geq8$ which implies $abc\geq1$ but $abc\leq1$. Where;s the mistake or the inequality is weak?

aayush-srivastava wrote: Hey can someone point out the mistake:- $abc(a+b+c)=3.$ Using $A.M.-G.M.$ we have $abc\leq1$ $(a+b)(b+c)(c+a)\geq 8.$ on expanding becomes $(2,1,0)+2abc\geq8$. As $(2,1,0)\succ(1,1,1)$ we get $8abc\geq8$ which implies $abc\geq1$ but $abc\leq1$. Where;s the mistake or the inequality is weak? I thought exactly the same, don't know

aayush-srivastava wrote: Hey can someone point out the mistake:- $abc(a+b+c)=3.$ Using $A.M.-G.M.$ we have $abc\leq1$ $(a+b)(b+c)(c+a)\geq 8.$ on expanding becomes $(2,1,0)+2abc\geq8$. As $(2,1,0)\succ(1,1,1)$ we get $8abc\geq8$ which implies $abc\geq1$ but $abc\leq1$. Where;s the mistake or the inequality is weak? There isn't any mistake, just turning something into $(1,1,1)$ usually makes it really weak.

Potla wrote: Let $a,b,c$ be positive real numbers such that $abc(a+b+c)=3.$ Prove that we have \[(a+b)(b+c)(c+a)\geq 8.\]Also determine the case of equality. We have \[(a+b)(b+c)(c+a)-8=\frac{1}{12}\sum{bc(b+c)(a^2+2a+9)(a-1)^2}+\frac{1}{6}abc(ab+ac+bc-3)^2\ge{0}.\]

It is clear that $3 \ge abc(a+b+c) \ge 3(abc)^{\frac{4}{3}}$ gives $abc \le 1$, so $$(a + b)(b + c)(c + a) - 8 = (a+b+c)(ab+bc+ca) - 8 - abc = \frac{3(ab+bc+ca)}{abc} - 8 - abc \ge \frac{9}{(abc)^\frac{1}{3}} - 8 - abc \ge 0.$$Equality in AM-GM occurs when all variables are equal, so $a = b = c = 1$.

Potla wrote: Let $a,b,c$ be positive real numbers such that $abc(a+b+c)=3.$ Prove that we have \[(a+b)(b+c)(c+a)\geq 8.\]Also determine the case of equality. Does the following solution work?

Thanks in advance.

anandiyer12 wrote: Potla wrote: Let $a,b,c$ be positive real numbers such that $abc(a+b+c)=3.$ Prove that we have \[(a+b)(b+c)(c+a)\geq 8.\]Also determine the case of equality. Does the following solution work?

Thanks in advance.

anandiyer12 wrote: Multiplying the three inequalities we have that $(a+b)(b+c)(c+a)\ge 8abc$ and we already know that $abc$ is at max $1$ so we are done. If $abc=\frac{1}{2}$, then all you proved is $(a+b)(b+c)(c+a)\geq 4$. Your inequality is weaker than the desired inequality. oosp sniped