<p>Prove: |a+b| / ( 1+|a+b| ) ≤ |a| / (1+|a| ) + |b| / ( 1+|b| )? Thanks!!!</p>
<p>a=1; b=1</p>
<p>2/3 < (1/2) + (1/2)
2/3 < 1</p>
<p>First of all you cant just use a scenario of a and b kieran to solve a proof…
anyway:
consider a+b=c
(|c|/1+|c|) will always be less than |b|/(1+|b|) + |a|/(1+|a|)
because for each value of a and b, the right side is bigger than or equal to 1 and the left side is less than one with the summation of a+b. Then of course if you plug in 0 for a and b it’s equal to 0 for that instance.</p>
<p>What if 0<a,b<1 ? Does anyone else have any ideas?</p>
<p>Like Raichu said, the BEST way to solve these kinds of questions on the SAT are to pick numbers that fit into the given parameters for a,b,c,etc. Then you see which options match your answer. This method takes the complexity out of the problem, turing a potentially difficult question, into an extremely easy one (computer using calculator for assurance).</p>
<p>It’s not an sat question. Does anyone know how to solve this proof?? I think it ha to do with concavity but im not sure</p>
<p>This is definitely not a SAT Reasoning question, nor a SAT II nor an AP question. So if you’re not focused on any of these areas skip this (and previous posts).</p>
<p>It’s actually a pretty hard problem. There isn’t a simple and quick answer for the general case. Perhaps you can re-post in a more suitable forum to get alternate proofs.</p>
<p>It’s very easy to prove for the case where a and b are both positive, or when a and b are both negative. Raichu (above) provided that proof. What’s not so easy (so far as I can see) is the subcase where a is positive and b negative, or conversely b positive and a negative. </p>
<p>Take the first subcase. Write b as a multiple of a – i.e. b = -xa. Consider first the case where x is between 0 and 1. I’ll gloss over the case where x is greater than 1.</p>
<p>Substitute b = -xa. Rewrite the desired inequality in terms of the function.</p>
<p>f(a,x) = a / (1+a) -ax / ( 1+ax ) - (a-ax) / ( 1+(a-ax))</p>
<p>If f(a,x) is always greater or equal to zero then we have the proof.</p>
<p>Note first that f(a,0) is 0. Take the derivative df(a,x)/dx. Note that it is positive in the range x=0+ and x=1. So if f(a,0) is 0 and the derivative is positive for all x between 0+ and 1 then f(a,x) is positive in the range x=0+ to 1. That proves the inequality for this subcase. Then repeat for the subcase where x is greater than 1. You’ll find that the function starts out as 0 or greater for x = 1 and that the derivative is positive for all x > 1.</p>
<p>If this was a SAT level question, ie choose the following inequalities that are always true, just take two values ( as in a=3 b= 5) and plug away until you get a correct response.</p>