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Tetrahedrons and spheres
ReticulatedPython   4
N Apr 28, 2025 by soryn
Let $OABC$ be a tetrahedron such that $\angle{AOB}=\angle{AOC}=\angle{BOC}=90^\circ.$ A sphere of radius $r$ is circumscribed about tetrahedron $OABC.$ Given that $OA=a$, $OB=b$, and $OC=c$, prove that $$r^2+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9\sqrt[3]{4}}{4}$$with equality at $a=b=c=\sqrt[3]{2}.$
4 replies
ReticulatedPython
Apr 21, 2025
soryn
Apr 28, 2025
Tetrahedrons and spheres
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ReticulatedPython
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#1 • 1 Y
Y by Exponent11
Let $OABC$ be a tetrahedron such that $\angle{AOB}=\angle{AOC}=\angle{BOC}=90^\circ.$ A sphere of radius $r$ is circumscribed about tetrahedron $OABC.$ Given that $OA=a$, $OB=b$, and $OC=c$, prove that $$r^2+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9\sqrt[3]{4}}{4}$$with equality at $a=b=c=\sqrt[3]{2}.$
This post has been edited 19 times. Last edited by ReticulatedPython, Apr 22, 2025, 5:31 PM
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jb2015007
1954 posts
#2
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bruh what the bash
lol
im doing this when i get home
looks nice
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ReticulatedPython
684 posts
#3
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Hint
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vanstraelen
9026 posts
#4
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The equation of the sphere $(x-\frac{a}{2})^{2}+(y-\frac{b}{2})^{2}+(z-\frac{c}{2})^{2}=\frac{1}{4}(a^{2}+b^{2}+c^{2})$.
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soryn
5342 posts
#5
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Yes, r^2=1/4•(a^2+b^2+c^2) and S=1/4•(a^2+b^2+c^2)+(1/a+ 1/b+1/c)=S1+S2. By AM-GM, S1>=3/4•(abc)^2/3 and S2>=3•(abc)^1/3. Now, denote (abc)^1/3=x, and study the variation of function f(x)=3/4•x^2+ 3/x. This leads to the desired result.
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