# 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.

Bạn đang xem: If abc then prove that cos a cos b cos c 1 2cosacosbcosc 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.

Xem thêm: Đồng Hồ Tự Thiết Kế Đồng Hồ Đeo Tay Tự Thiết Kế, Tự Thiết Kế Đồng Hồ

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.

nói qua
Cite
Follow
edited Mar 23, 2017 at 15:32
answered Jan 11, 2015 at 21:56 user133281user133281
\$endgroup\$
địa chỉ cửa hàng a phản hồi |
1
\$egingroup\$
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\$.

Xem thêm: Tạo File Gõ Tắt Trong Unikey, Cách Sử Dụng Tính Năng Gõ Tắt Trên Unikey

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.