In an acute-angled triangle ABC, AB < AC, BD and CE are the altitudes. Prove that $AB^2 + CE^2 < AC^2 + BD^2$


$AB^n + CE^n < AC^n + BD^n$ for positive integers n

I have already found BD < CE and AD < AE but cannot seem to find more information. I already tried using the pythagorean and triangle areas to try and prove it.


We have $$A=\frac{ch_c}{2}$$ and $$A=\frac {bh_b}{2}$$ so we have to Show that

$$b^2+\frac{4A^2}{b^2}<b^2+\frac{4A^2}{c^2}$$ this is equivalent to

$$(b^2-c^2)(1-\frac{4A^2}{b^2c^2})>0$$ this is true since we have $$c<b$$ and $$A=\frac{1}{2}bc\sin(\gamma)\le \frac{bc}{2}$$ and your second inequality is equivalent to

$$(b^n-c^n)\left(1-\frac{(2A)^n}{(bc)^n}\right)>0$$ the rest is for you!


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