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I am asked to "prove or disprove the following statement using contradiction" --> If a, b, c are the sides of a right traingle and c is the hypotenuse then $c<a+b$. Here is my proposed proof:

Proof: Assume that $a,b,c$ are sides of a right traingle and $c$ is the hypotenuse, and $c\geq a+b$. Then $c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}$. Substituting, we have $\sqrt{a^2+b^2}\geq a+b\implies a^2+b^2\geq a^2+b^2+2ab$. This is clearly a contradiction, so we have proven the original statement. $\blacksquare$

Please let me know if I've made any incorrect statements or left out any neccessary rigor.

Brandon O'Neil
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1 Answers1

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I don't know it is correct or not but here is a sort deduction of your problem. Let us consider a triangle which is not Right angled (shown in figure)

Here the red line is the perpendicular dropped from the top, to the line of length $c$, is of length $h>0$. Now, as you proved the inequality for right triangle we get, $a<h+\frac{c}{2}-x$ and $b<h+\frac{c}{2}+x$ where $x\in [0,\frac{c}{2}]$. Now, $a+b<2h+c\iff a+b-c<2h$. So we must have $a+b-c>0$ if not then, $a+b-c\le 0$. Now if the equality holds then no triangle wil exists, else if $a+b-c<0$ then $\frac{a+b}{c}<1$ again from Pythagoras theorem, $h^2+(\frac{c}{2}-x)^2=a^2$ and $h^2+(\frac{c}{2}+x)^2=b^2$. Solving for $x$ we get, $x=\frac{b^2-a^2}{2c}=\frac{a+b}{c}\frac{b-a}{2}<\frac{b-a}{2}$ since, $\frac{a+b}{c}<1$, and this is true for all $x\in [0,\frac{c}{2}]$ so, $\frac{c}{2}<\frac{b-a}{2}$ implies that $a-b+c<0$ similarly $-a+b+c<0$ adding up we get $0<0$ contradiction!! So $a+b-c>0$. Proved! Please correct my proof if necessary. Thank you.

Sujit Bhattacharyya
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