Inequality. (1 Viewer)
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Carrotsticks
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I had a rather different approach, but good solution!
We are given that:
(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})
We are given that:
seanieg89
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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.
Good solution carrotsticks, essentially the same as mine.
RealiseNothing
what is that?It is Cowpea
I'm not going to prove the basic results (AM>= GM) but here it is:
^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*})
Doesn't imply that:
seanieg89
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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:
\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.)