Given thata+b=c+d=10 and |a−b|&gt;|c−d|, show that ab&lt;cd

Step 1Alternative approacha,b=5+r, 5−r in some order.c,d=5+s, 5−s in some order.Since |a−b|&gt;|c−d|, you have that r&gt;sStep 2See comments following answer. Above should be|r|&gt;|s|, or I should have specified that r,s were positive.Therefore (ab)=52−r2&lt;52−s2=(cd)

Step 1|a−b|≥|c−d|→|a−b|2≥|c−d|2→(a+b)2−4ab≥(c+d)2−4cd

Step 1Alternative approacha,b=5+r, 5−r in some order.c,d=5+s, 5−s in some order.Since |a−b|&gt;|c−d|, you have that r&gt;sStep 2See comments following answer. Above should be|r|&gt;|s|, or I should have specified that r,s were positive.Therefore (ab)=52−r2&lt;52−s2=(cd)

Step 1|a−b|≥|c−d|→|a−b|2≥|c−d|2→(a+b)2−4ab≥(c+d)2−4cd

Best solutions to - "Given thata+b=c+d=10 and |a−b|>|c−d|, show that ab<cd"

High School
Algebra
Algebra I
Quadratic function and equation

Petrolovujhm

Answered question

2022-03-24

Given that

a + b = c + d = 10

and

| a - b | > | c - d |

, show that

a b < c d

Answer & Explanation

SofZookywookeoybd

Beginner2022-03-25Added 15 answers

Step 1
Alternative approach

a, b = 5 + r, 5 - r

in some order.

c, d = 5 + s, 5 - s

in some order.
Since

| a - b | > | c - d |

, you have that

r > s

Step 2
See comments following answer. Above should be

| r | > | s |

, or I should have specified that r,s were positive.
Therefore

(a b) = 5^{2} - r^{2} < 5^{2} - s^{2} = (c d)

Melody Gamble

Beginner2022-03-26Added 10 answers

Step 1

| a - b | \geq | c - d |

\to {| a - b |}^{2} \geq {| c - d |}^{2}

\to {(a + b)}^{2} - 4 a b \geq {(c + d)}^{2} - 4 c d

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