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Does ${a^2+b^2}\gt{c^2+d^2}$ imply ${a+b}\gt{c+d}$ where $a, b, c,$ and $d$ are positive? Is the converse also true? I have not found any exceptions to the rule by randomly substituting values in the inequality, and neither have I found a proof for the same.

(Edit)
Ight, context: I had to prove that the law of reflection of light results in the shortest path:
Ray Diagram, with constructions and all

Here's what I did:
In triangles API, BQI, API' and BQI' respectively: $${AI}^2 = {AP}^2 + {IP}^2 \dots (i)$$ $${BI}^2 = {BQ}^2 + {IQ}^2 \dots (ii)$$ $${AI'}^2 = {AP}^2 + {I'P}^2 \dots (iii)$$ $${BI'}^2 = {BQ}^2 + {I'Q}^2 \dots (iv)$$

Adding $(i)$ and $(ii)$: $${AI}^2 +{BI}^2 = {AP}^2 + {BQ}^2 + {IP}^2 + {IQ}^2 \dots (a)$$ Adding $(iii)$ and $(iv)$: $${AI'}^2 +{BI'}^2 = {AP}^2 + {BQ}^2 + {I'P}^2 + {I'Q}^2 \dots (b)$$

Also: $$ \begin{align} {I'P}^2 + {I'Q}^2 &= (IP - II')^2 + (IQ + II')^2 \\ &= {IP}^2 + {IQ}^2 + 2II'(IQ-IP+II') \dots (c) \\ \end{align} $$

Therefore: $$ \begin{align} &IQ \gt IP \\ \implies &IQ - IP \gt 0 \\ \implies &IQ-IP+II' \gt 0 \\ \implies &{IP}^2 + {IQ}^2 + 2II'(IQ-IP+II') \gt {IP}^2 + {IQ}^2 \\ \implies &{I'P}^2 + {I'Q}^2 \gt {IP}^2 + {IQ}^2 \,\,\,\,\, [from \, (c)]\\ \implies &{AP}^2 + {BQ}^2 + {I'P}^2 + {I'Q}^2 \gt {AP}^2 + {BQ}^2 + {IP}^2 + {IQ}^2 \\ \implies &{AI'}^2 +{BI'}^2 \gt {AI}^2 + {BI}^2 \,\,\,\,\, [from \, (a), \, and \, (b)]\\ \implies & AI' + BI' \gt AI + BI \end{align} $$

In the comments, my abstracted version of the last step has been proved incorrect through counterexamples, but it does make sense in the situation. Of, couse no amount of head banging will justify my answer over the actual one, but I am still curious if this can be saved, and what I might be missing here.

Annoying Cat
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  • $6^2 + 4^2 > 5^2 + 5^2$, but $6 + 4 = 5 + 5$. – N. F. Taussig May 30 '23 at 11:53
  • See also https://math.stackexchange.com/q/3400769/42969 and https://math.stackexchange.com/q/3472098/42969 for counterexamples. – Martin R May 30 '23 at 11:53
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    $0^2+10^2 > 5^2+6^2$ but $0+10<5+6$. – Surb May 30 '23 at 11:54
  • Check [[ https://math.stackexchange.com/questions/4699465/is-the-taxicab-metric-equivalent-to-the-euclidean-metric/4699476#4699476 ]] to get more terminology. – Prem May 30 '23 at 11:55
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    $1^2+5^2>3^2+4^2$, but $1+5<3+4$. You did not look very far, did you? – José Carlos Santos May 30 '23 at 12:01
  • No, but it does imply $(a-d)(a+d) \gt (c-b)(c+b)$. For example, if $a = 3, b = 6, c=d=4$, then $a^2+b^2 = 9+36=45$ and $c^2+d^2=16+16=32$, and $(a-d)(a+d) = -7$, while $(c-b)(c+b) = -20$. $45 \gt 32$ and $-7 \gt -20.$ – H. sapiens rex May 30 '23 at 12:14
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    I noticed that there is no place in your proof where you use the premise that $\angle AIP=\angle BIQ$ or any equivalent statement. This should be a huge red flag. Also notice that you looked at a point $I'$ in the smaller of the two segments $IP$ and $IQ$ but never in the larger segment. In fact, if the last step in your proof were valid, it would imply that the shortest path from $A$ to the line $PQ$ to $B$ is via the midpoint of $PQ$, which is not true in general. – David K Jun 01 '23 at 18:49
  • @DavidK The answer that was given to me (in school) involves reflecting triangles $AIP$ and $AI'P$ below $PQ$ and using the Triangle Inequality: https://imgur.com/a/rk5G4V6 Notice how this proof also does not use the (law of reflection)[https://en.wikipedia.org/wiki/Specular_reflection]... or so I thought. For $A'BI'$ to be a triangle, it is necessary to prove that A'B is a straight line, which requires that $\angle AIP=\angle BIQ$. This realisation addresses all my confusion pretty well, so thank you :) – Annoying Cat Jul 04 '23 at 09:58
  • No idea what you mean by your next sentence though. I don't see how $AI' + BI' \gt AI + BI$ implies anything about $I$ being the midpoint of $PQ$. The diagram is a little off in that regard. That's an error on my part. And so is not reading my notifications properly. Sorry for the delay. – Annoying Cat Jul 04 '23 at 10:05
  • It’s always a bit weird to say “If this false thing were true then this other thing would be true.” But I think what I was thinking was that if it were true that $${AI'}^2 +{BI'}^2 \gt {AI}^2 + {BI}^2 \stackrel?\implies AI' + BI' \gt AI + BI,$$ then the way to minimize $AI+BI$ would be to minimize $AI^2+BI^2$, which (by (a)) is equivalent to minimizing $IP^2+IQ^2$, which happens when $I$ is the midpoint of $PQ$. – David K Jul 04 '23 at 22:32
  • @Davidk I see.. – Annoying Cat Jul 05 '23 at 14:20

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Evidently from the comments we have counterexamples. However with all variables positive,

$a^2+b^2>c^2+d^2\implies\sqrt2(a+b)>c+d.$

Oscar Lanzi
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