Ineq-Ligouras-G006 - Geometry - Old & easy

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Ineq-Ligouras-G006 - Geometry - Old & easy

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#1 h Jan 5, 2011, 10:22 PM • 2 Y

Y by Adventure10, Mango247

Let $a, b$ and $c$ be the lengths of the sides of a triangle $ABC$ . If $a+b+c = 1$ , prove that

${{1}\over{a^2}}+{{1}\over{b^2}}+{{1}\over{c^2}}+{{1}\over{4abc}}\geq \ 32$

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Victoria_Discalceata1

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Victoria_Discalceata1

#2 h Jan 31, 2020, 1:45 PM • 1 Y

Y by Adventure10

In terms of $R,r,s$ (circumradius, inradius, semiperimeter) we have $\sum a=2s,\ \sum ab=4Rr+r^2+s^2,\ abc=4Rrs$ and then it suffices to prove $s^4+2r(r-3R)s^2-r^2\left(112R^2-8Rr-r^2\right)\ge 0\qquad (*)$ Since by Gerretsen and Euler $2s^2-2r(r-3R)\ge 2\left(16Rr-5r^2\right)-2r(r-3R)=r(13R-4r)\ge 11Rr\ge 0$ hence the LHS of $(*)$ is increasing in $s^2$ and it only suffices to check $s^2=16Rr-5r^2$ which yields $2r^2\left(24R^2-45Rr+8r^2\right)\ge 2r^3(3R+8r)\ge 0$ as desired.

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#3 h Jan 31, 2020, 2:21 PM • 1 Y

Y by Adventure10

When did it get the equal sign? Thanks.

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Victoria_Discalceata1

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Victoria_Discalceata1

#4 h Jan 31, 2020, 3:32 PM • 1 Y

Y by Adventure10

We do not have equality here. I can prove that the RHS of the original inequality can be increased to $\frac{135}{4}$ . In that case equality is attained for equilateral triangles.

This post has been edited 1 time. Last edited by Victoria_Discalceata1, Jan 31, 2020, 3:42 PM

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