How can this inequality be proven for positive values of a, b, c, and d?

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The inequality $\sqrt {b^2+c^2}+\sqrt {a^2+c^2+d^2+2cd}>\sqrt {a^2+b^2+d^2+2ab}$ is proven for positive values of $a$, $b$, $c$, and $d$. The discussion emphasizes a geometric approach to demonstrate the validity of the inequality, highlighting its elegance. Participants agree on the effectiveness of this method, confirming its applicability in mathematical proofs.

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prove the following

$a,b,c,d>0$

prove :$\sqrt {b^2+c^2}+\sqrt {a^2+c^2+d^2+2cd}>\sqrt {a^2+b^2+d^2+2ab}$
 
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My solution:

\[\sqrt{b^2+c^2}+\sqrt{a^2+(c+d)^2} > \sqrt{(a+b)^2+d^2}\\\\ b^2+c^2+a^2+(c+d)^2 +2\sqrt{b^2+c^2}\sqrt{a^2+(c+d)^2}> (a+b)^2+d^2\\\\ 2\sqrt{b^2+c^2}\sqrt{a^2+(c+d)^2}> (a+b)^2+d^2-a^2-b^2-c^2-(c+d)^2 \\\\ \sqrt{a^2b^2 +(b^2+c^2)(c+d)^2+a^2c^2} > ab-c^2-cd\]

The LHS $ > ab$, and the RHS $< ab$, because all four variables are positive reals. Thus the stated inequality holds.
 
Albert said:
prove the following

$a,b,c,d>0$

prove :$\sqrt {b^2+c^2}+\sqrt {a^2+c^2+d^2+2cd}>\sqrt {a^2+b^2+d^2+2ab}---(1)$
using geometry:
construct a rectangle with lengths $a+b,$ and $c+d$
Can you see ?$(1)$ automatically holds
 
I´m afraid not :( - please explain
 
lfdahl said:
I´m afraid not :( - please explain
ABCD is a rectangle ,CD=c+d=CQ+QD
BC=a+b=BP+PC
we have AP+PQ>AQ
by pythagorean theorem , and (1) holds
 
Albert said:
ABCD is a rectangle ,CD=c+d=CQ+QD
BC=a+b=BP+PC
we have AP+PQ>AQ
by pythagorean theorem , and (1) holds

Yes, of course! Very elegant! (Yes)
 

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