In an equilateral triangle, prove that $|BQ| + |PQ| + |CP| > 2l$

Upon reflection ...

enter image description here

$$2s = |\overline{AB^\prime}| \leq |\overline{AP}|+|\overline{PQ^\prime}|+|\overline{Q^\prime B^\prime}| = p+q+r$$


Let $AQ=x$, $AP=y$ and $l=1$.

Thus, $$PQ=\sqrt{x^2-xy+y^2},$$ $$PC=\sqrt{y^2-y+1}$$ and $$BQ=\sqrt{x^2-x+1}$$ and we need to prove that $$\sqrt{x^2-xy+y^2}+\sqrt{x^2-x+1}+\sqrt{y^2-y+1}\geq2.$$ Now, by Minkowwski $$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}+\sqrt{\left(y-\frac{1}{2}\right)^2+\frac{3}{4}}\geq$$ $$\geq\sqrt{\left(x-\frac{1}{2}+y-\frac{1}{2}\right)^2+\left(\frac{\sqrt3}{2}+\frac{\sqrt3}{2}\right)^2}=\sqrt{(x+y-1)^2+3}$$ and $$\sqrt{x^2-xy+y^2}\geq\frac{x+y}{2}.$$ Let $x+y=2a$.

Hence, $a\leq1$ and we need to prove that $$a+\sqrt{(2a-1)^2+3}\geq2$$ or $$\sqrt{4a^2-4a+4}\geq2-a$$ or $$4a^2-4a+4\geq a^2-4a+4,$$ which is obvious.

Done!

By my solution easy to make a geometric proof.

Tags:

Geometry

Inequality

Triangles

Geometric Inequalities

Related

$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}?$ Area of Intersection of Three Circles Fundamental Theorem of Calculus and discontinuous functions What is the intuition for using sigma algebras to define probability spaces? Proof of the Jordan Holder theorem from Serge Lang When all of $n-2$, $n+2$, $n^2-2$, and $n^2+2$ are primes? Can one recover $G$ from $G/H$? show that $\lim_{n \to \infty} n\int_0^1 e^{-tn}(\frac{\sinh t}{t})^n \, dt=1$ (Reference Request) Calculus on Banach Spaces Is this correct ${2\over 3}\cdot{6\over 5}\cdot{10\over 11}\cdots{4n+2\over 4n+2+(-1)^n}\cdots=\sqrt{2-\sqrt{2}}?$ How to look at the Lie derivative as a partial differential operator? How to find the limit of $\frac{\sqrt{x^2+9}-3}{x^2}$? as x approaches 0

Recent Posts