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.
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Tags: inequalities, geometry, circumcircle, trigonometry, geometric inequality, area of a triangle, Heron's formula
02.12.2012 19:38
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.
02.12.2012 19:46
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 $a=b=c=1.\Box$
03.12.2012 07:31
$3 = abc(a+b+c) \ge abc \cdot 3 \sqrt[3]{abc} = 3(abc)^{\frac43} \Rightarrow abc \le 1$ $\Rightarrow -abc \ge 1$ 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 $a=b=c=1$.
04.12.2012 14:15
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
04.12.2012 19:51
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
07.12.2012 03:18
$(a+b)(b+c)(c+a) = (a+b)(ab+(a+b+c)c)$ $= (a+b)(ab+\frac1{ab}+\frac1{ab}+\frac1{ab})\ge2\sqrt{ab}\cdot 4\sqrt[4]{\frac1{a^2b^2}}=8$.
07.12.2012 09:48
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
09.12.2012 07:56
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.
11.12.2012 01:57
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.
16.12.2012 18:54
this is equivalent to $\sin A\sin B\sin C\le\frac{3\sqrt{3}}{8}$
17.12.2012 00:50
sankha012 wrote: this is equivalent to $\sin A\sin B\sin C\le\frac{3\sqrt{3}}{8}$ yeah. $sinA=\frac{2\sqrt{xyz(x+y+z)}}{(z+x)(x+y)}.$
23.12.2012 03:00
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?
11.10.2013 15:25
Best way to be good at inequalities is to remember some very useful results.
12.10.2013 19:08
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 $a=b=c=1$.
12.10.2013 19:19
please hide solutions!
23.11.2015 17:26
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
23.11.2015 20:55
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?
23.11.2015 21:01
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
23.11.2015 21:09
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.
26.11.2015 02:29
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 \[(a+b)(b+c)(c+a)-8=\frac{1}{12}\sum{bc(b+c)(a^2+2a+9)(a-1)^2}+\frac{1}{6}abc(ab+ac+bc-3)^2\ge{0}.\]
26.11.2015 04:51
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 $a = b = c = 1$.
26.11.2015 05:02
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.
26.11.2015 05:14
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.
26.11.2015 05:15
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