Inequality. (1 Viewer)

asianese · May 4, 2012

I'm not going to prove the basic results (AM>= GM) but here it is:

$\begin{align*} a+b+c&\geq 3 \sqrt[3]{abc}\\ \text{Also} \,a^2+b^2+c^2 &\geq ab+bc+ca\\ \\ \text{Now}\, a^2+b^2+c^2 &\geq 3 \sqrt[3]{(abc)^2} \\ \text{Now from}\, a+b \geq 2\sqrt{ab}\, \text{etc. - Multiply to get} (a+b)(a+c)(b+c)&\geq 8abc = 1\\ \therefore abc=\frac{1}{8}\\ \\ \text{Now}\, a^2+b^2+c^2 &\geq 3 \sqrt[3]{(abc)^2}\\ &\geq 3\sqrt[3]{\left( \frac{1}{8} \right)^2} &= \frac{3}{4}\\ \therefore \frac{3}{4} \geq ab+bc+ca \\\text{As}\, a^2+b^2+c^2 \geq ab+bc+ca \end{align*}$

Carrotsticks · May 4, 2012

I had a rather different approach, but good solution!

We are given that:

$\\ (a+b)(a+c)(b+c) = 1 \\\\ (a+b+c -c)(a+b+c-b)(a+b+c-a) = 1 \\\\ $Using cyclic expansion, we easily acquire:$ \\\\ ab+ac+bc = \frac{1+abc}{a+b+c} \\\\ $Similarly to the above proof, we have $ abc \leq \frac{1}{8} $ and $ \frac{1}{a+b+c} \leq \frac{2}{3} \\\\ $Putting this into our identity yields the result directly:$ \\\\ ab+bc+ac \leq \left ( 1+\frac{1}{8}} \right ) \times \frac{2}{3} = \frac{3}{4}$

seanieg89 · May 5, 2012

Asianese, your solution doesn't quite work. There are a few inequality signs the wrong way around which ruin your idea for proving the inequality. It can be fixed but the entire second half of your argument will look different.

Good solution carrotsticks, essentially the same as mine.

asianese · May 5, 2012

Hmm I think I see it...it's where I subbed in $abc=\frac{1}{8}$ But I subbed a $\leq$ into a $\geq$ didn't I? I will re do it later.

RealiseNothing · May 5, 2012

asianese said:
I'm not going to prove the basic results (AM>= GM) but here it is:

$\begin{align*} a+b+c&\geq 3 \sqrt[3]{abc}\\ \text{Also} \,a^2+b^2+c^2 &\geq ab+bc+ca\\ \\ \text{Now}\, a^2+b^2+c^2 &\geq 3 \sqrt[3]{(abc)^2} \\ \text{Now from}\, a+b \geq 2\sqrt{ab}\, \text{etc. - Multiply to get} (a+b)(a+c)(b+c)&\geq 8abc = 1\\ \therefore abc=\frac{1}{8}\\ \\ \text{Now}\, a^2+b^2+c^2 &\geq 3 \sqrt[3]{(abc)^2}\\ &\geq 3\sqrt[3]{\left( \frac{1}{8} \right)^2} &= \frac{3}{4}\\ \therefore \frac{3}{4} \geq ab+bc+ca \\\text{As}\, a^2+b^2+c^2 \geq ab+bc+ca \end{align*}$

$a^2 + b^2 + c^2 \geq ab + bc + ca ~and~ a^2 + b^2 + c^2 \geq \frac{3}{4}$

Doesn't imply that:

$\frac{3}{4} \geq ab +bc + ca$

asianese · May 5, 2012

Hmm...I can't seem to get it to work using AM>= GM and other results. Is the cyclic factorisation the only way? (Probably not?)

seanieg89 · May 5, 2012

No, AM-GM is the only essential step really. Spotting a neat factorisation isn't entirely necessary, althought it makes the solution look nicer. One could also do the question like this, just using AM-GM and grouping suitable terms together:

$(a+b+c)\geq 3/2, abc\leq 1/8$ by AM-GM$.\\$So $1=(a+b)(a+c)(b+c)=ab(a+b)+ac(a+c)+bc(b+c)+2abc\geq ab(3/2-c)+ac(3/2-b)+bc(3/2-a)+2abc=(3/2)\cdot(ab+ac+bc)-abc.\\$Hence: $ab+ac+bc\leq (2/3)\cdot(1+abc)\leq(2/3)\cdot (9/8)=3/4.$

asianese · May 5, 2012

OH I see...thanks!!

Inequality. (1 Viewer)

seanieg89

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asianese

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Carrotsticks

Retired

seanieg89

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asianese

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RealiseNothing

what is that?It is Cowpea

asianese

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seanieg89

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asianese

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