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Exercise 9 · Q6

Q.Prove that the following inequality holds true: 5+3>6+2\sqrt{5} + \sqrt{3} > \sqrt{6} + \sqrt{2}

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Both sides are positive, so comparing their squares 8+2158+2\sqrt{15} and 8+2128+2\sqrt{12} settles it: the left is larger, hence 5+3>6+2\sqrt{5}+\sqrt{3}>\sqrt{6}+\sqrt{2}.

For positive numbers P,Q>0P,Q>0: P>Q  ⟺  P2>Q2P > Q \iff P^2 > Q^2. Use (x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2.

  1. Let P=5+3P = \sqrt{5}+\sqrt{3} and Q=6+2Q = \sqrt{6}+\sqrt{2}; both are positive.
  2. Square PP:

P2=(5+3)2=5+3+25⋅3=8+215P^2 = (\sqrt5+\sqrt3)^2 = 5 + 3 + 2\sqrt{5\cdot3} = 8 + 2\sqrt{15}

  1. Square QQ:

Q2=(6+2)2=6+2+26⋅2=8+212Q^2 = (\sqrt6+\sqrt2)^2 = 6 + 2 + 2\sqrt{6\cdot2} = 8 + 2\sqrt{12}

  1. Compare squares: both have the constant 88, so compare 2152\sqrt{15} with 2122\sqrt{12}. Since 15>1215 > 12, 15>12\sqrt{15} > \sqrt{12}, hence …

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