MHB 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 to be proven for positive values of a, b, c, and d. A geometric approach is suggested as an elegant method for the proof. Participants express agreement on the effectiveness of this geometric perspective. The discussion emphasizes the importance of maintaining positivity in the variables involved. Overall, the focus is on finding a clear and rigorous proof of the stated inequality.
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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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