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Suppose $\triangle ABC$ is an obtuse triangle with side lengths $a=BC, b=CA, c=AB$. I want to show that $$a^3\cos A+b^3\cos B+c^3\cos C<abc.$$

My idea is to use the cosine rule. I have $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, etc. Plugging into the inequality I get

$$a^4b^2+a^4c^2-a^6+b^4a^2+b^4c^2-b^6+c^4a^2+c^4b^2-c^6<2a^2b^2c^2.$$

How can I show this?

Kunal
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  • You have an obtuse triangle. What can you say about the lengths of the sides? Choose one of them to be opposite the obtuse angle. (I imagine that the fact that the triangle is obtuse is what drives this inequality.) – John Nov 12 '13 at 18:55

2 Answers2

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Using $\sin2x,\cos2x$ formula

$$a^3\cos A=(2R\sin A)^3\cos A= 2R^3(2\sin^2A)(2\sin A\cos A)$$ $$=2R^3(1-\cos2A)\sin2A=R^3(2\sin2A-2\sin2A\cos2A)=R^3(2\sin2A-\sin4A)$$

Using this, $\sum \sin2A=4\prod \sin A$

Now, $\sin4A+\sin4B+\sin4C=2\sin(2A+2B)\cos(2A-2B)+2\sin2C\cos2C$

Now, $\cos2C=\cos\{2\pi-2(A+B)\}=\cos2(A+B)$ and $\sin2(A+B)=\sin(2\pi-2C0=-\sin2C$

$\implies\sin4A+\sin4B+\sin4C=-2\sin2C\cos(2A-2B)+2\sin2C\cos2(A+B)$ $=-\sin2C\{\cos(2A-2B)-\cos(2A+2B)\}=-\sin2C\cdot2\sin2A\sin2B$ $=-2(2\sin C\cos C)(2\sin A\cos A)(2\sin B\cos B)$

For an obtuse triangle, only one angle is between $(\frac\pi2,\pi)$ so exactly one of the cosine ratio $<0$ and all the sine ratios are $>0$

$\implies\sin4A+\sin4B+\sin4C>0$

$\implies\sum a^3\cos A <2R^3(\sum\sin2A)=2R^3(4\sin A\sin B\sin C)=\prod(2R\sin A)$

Community
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lab bhattacharjee
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  • This looks good, but I still wonder if there's an easier way. – Kunal Nov 12 '13 at 19:55
  • @Kunal, there are two aspects of a problem : $1):$ solution using legitimate formulae and $2):$ solution based on how the problem came to being. My method definitely qualifies $1),$ I am not sure of $2)$ – lab bhattacharjee Nov 13 '13 at 03:52
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here is another way follow op's idea:

for easy, $x=a^2,y=b^2,z=c^2,$ WOLG, let $C$ is obtuse triangle, then $ x+y<z$, we want to prove :

$x^2y+x^2z-x^3+y^2x+y^2z-y^3+z^2x+z^2y-z^3<2xyz$

that is to prove: when $z>x+y$

$f(z)=-(x-y)^2(x+y)+(x-y)^2z+(x+y)z^2-z^3 <0$

now we prove $f(z)$ is mono decreasing function:

$f'(z)=(x-y)^2+2(x+y)z-3z^2$

$f''(z)=2(x+y)-6z<0 \implies f'(z)<f'(x+y)=-4xy<0 \implies f(z) <f(x+y)=0$

chenbai
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