Question

Quick Algebra help
Will fan and medal
question to come

Answer 1

\[a^2+b^2=2h^2+d^2+e^2\]
\[d+e=c\]
Prove \[a+b < c+h\]

Answer 2

@ganeshie8
@ikram002p

Answer 3

Well it appears to me

\[0 < 2h^2+d^2+e^2, a,b \in \mathbb{R} \]

Answer 4

can you guys make it a litte easire? im only a high school junior

Answer 5

For the real numbers, a square must be positive, so the sum of two squares must be a positive number, so something equal to the sum of two squares must be greater than zero.

Answer 6

yea i get that i just cant simplify the simultaneous equation to solve for the final target

Answer 7

\[d+e=c \iff e = c - d \]

so

\[e^2 = (c-d)^2 = c^2 -2cd + d^2 \]

Answer 8

oops.

\[d^2 + e^2= c^2 - 2cd + 2d^2\]

Answer 9

ok then

Answer 10

\[0 < 2h^2+ c^2 - 2cd + 2d^2\]

Answer 11

could you write our a procedure? @LarsEighner because i have some other work and will come back later thanks

Answer 12

@LarsEighner is there any quick way to do this?

Answer 13

Hm. I think we want to go back.

\[a^2+b^2=2h^2+d^2+e^2\]

\[ e^2 = (c-d)^2\]
\[d^2 = (c - e)^2\]

\[a^2+b^2=2h^2+(c^2 -2ce + e^2)+(c^2 -2cd + d^2)\]

\[a^2+b^2=2h^2+ d^2 + e^2 + (c^2 -2ce)+(c^2 -2cd)\]

I'm not finding one.

Answer 14

ok thanks for your help!

Answer 15

\[\large

\begin{align}
(a+b)^2 &= a^2+b^2+2ab \\
&= 2h^2+d^2+e^2+2ab \\
&= 2h^2 + (d+e)^2 + 2ab-2de \\
&= 2h^2 + c^2 + 2ab-2de\\
&= (h+c)^2 +h^2+ 2ab - 2de-2hc

\end{align}

\]

Answer 16

from where you got this problem ?

Answer 17

from some test paper

Answer 18

cool :D