If abc then prove that cos a cos b cos c 1 2cosacosbcosc


I"m looking for an elementary, geometric proof. So avoid derivatives and tools from optimalization theory if it"s possible.

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Writing $cos 2x = 1 -sin^2 x$ & using the sine rule $fracasin alpha = 2R$, the inequality turns out to lớn be equivalent to$$ a^2 + b^2 + c^2 leq 9R^2 $$where $a,b,c$ are the sides of the triangle & $R$ is the circumradius.

This inequality is sometimes known as Leibniz"s inequality.

For this inequality, a geometric proof is possible.Let $O$ and $G$ be the circumcenter và centroid of the triangle, respectively. Let $A$ be one of the vertices and let $M$ be the midpoint of the side $BC$ opposite khổng lồ $A$. Applying Stewart"s theorem in triangles $ABC$ và $AOM$, one can show that $$|OG|^2 = R^2 - frac19(a^2+b^2+c^2), qquad (ast)$$thus proving the inequality.

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An equivalent size of $(ast)$ is $|OH|^2 = 9R^2 - a^2-b^2-c^2$, where $H$ is the orthocenter. This equality can be proven using complex numbers and surely also by purely geometric means, for instance by computing the power nguồn of $H$ with respect khổng lồ the circumcircle.

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edited Mar 23, 2017 at 15:32
answered Jan 11, 2015 at 21:56

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If you are interested in an algebraic proof, mine goes as follows:

$cos2A + cos2B = 2 cos(A+B) cos(A-B) = %2 cos(pi - C) cos(A-B) = \-2 cosC cos(A-B)$

Hence,egineqnarray*& cos2A + cos2B + cos2C &geq& - dfrac32 \iff và -2 cosC cos(A-B) + cos2C &geq& - dfrac32 \iff & -2 cosC cos(A-B) + 2cos^2C - 1 &geq& - dfrac32 \iff và 4cos^2C - 4cos(A-B) cosC + 1 &geq& 0endeqnarray*

We prove the last inequality with a quadratic function.

Consider $f(t) = 4t^2 - 4 cosphi ; t + 1$, $t, phi in aquabigman.combbR$. The descriminant is given by $D = 16(cos^2phi - 1) leq 0$. Hence $f(t) geq 0$ for all $t, phi in aquabigman.combbR$. The result is now obvious if we let $t = cosC$ và $phi = A - B$.

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Also, $f(t) = 0$ if và only if $phi = npi$, $n in aquabigman.combbZ$, which is translated to $A = B$ for triangle $ABC$. In this case, $t = 1/2$, which means that $cosC = 1/2 implies C = fracpi3$. Thus, equality holds if & only if $A = B = C = fracpi3$. I.e. If và only if the triangle is equilateral.