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

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Discussion Overview

The discussion revolves around proving an inequality involving positive values of variables a, b, c, and d. The scope includes mathematical reasoning and potentially geometric interpretations related to the inequality.

Discussion Character

  • Mathematical reasoning, Exploratory, Technical explanation

Main Points Raised

  • One participant presents the inequality to be proven, stating the conditions for a, b, c, and d as positive values.
  • Another participant offers a solution approach, indicating a geometric method may be used to prove the inequality.
  • A subsequent reply expresses approval of the proposed solution, highlighting its elegance.

Areas of Agreement / Disagreement

Participants appear to agree on the validity of the proposed geometric approach, but the discussion does not resolve whether the inequality is proven or not.

Contextual Notes

The discussion does not provide detailed assumptions or definitions that may affect the proof, nor does it clarify any unresolved mathematical steps.

Albert1
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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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