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CDS Mathematics · Algebraic Identities and Simplification

Sums of Squares and Least Values

A square is never negative, so a sum of squares that equals zero forces each square to be zero, and t + 1/t is never less than 2 for positive t.

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Why this matters

Twelve PYQs, five of them HARD, and many are data-sufficiency items. One fact runs the whole page: a square is never negative. It decides when numbers must be equal, gives the least value of expressions like a + 1/a, and settles most 'which is larger' comparisons.

Concept 1 of 3: Sums of squares that vanish

Squares cannot cancel each other because none of them is negative. So if a sum of squares is zero, each square is zero. Complete the squares first, then read off the values.

Definition

  • If p2+q2+⋯=0p^2 + q^2 + \cdots = 0 for real numbers, then p=q=⋯=0p = q = \cdots = 0.
  • a2+b2+c2−ab−bc−ca=12[(a−b)2+(b−c)2+(c−a)2]≥0a^2 + b^2 + c^2 - ab - bc - ca = \dfrac12\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right] \ge 0, and it is 00 only when a=b=ca = b = c.
  • x2+y2−2xy=(x−y)2x^2 + y^2 - 2xy = (x - y)^2, so xy+yx=2\dfrac xy + \dfrac yx = 2 forces x=yx = y.
  • An expression that looks lopsided can still be a sum of squares: expand a guess like (a−b)2+k(a−c)2+k(b−c)2(a - b)^2 + k(a - c)^2 + k(b - c)^2 and compare.

Sum of squared differences

a2+b2+c2−ab−bc−ca=12[(a−b)2+(b−c)2+(c−a)2]a^2 + b^2 + c^2 - ab - bc - ca = \tfrac12\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right]

Worked example

If x2+y2−4x+6y+13=0x^2 + y^2 - 4x + 6y + 13 = 0, find xx and yy.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2023 · CDS (II) 2023 — Elementary Mathematics · Q83Hard

Example 1 · Algebraic Identities and Simplification · Sums of Squares and Least Values

If a,b,c,x,y,za, b, c, x, y, z are real numbers such that (a+b+c)2−3(ab+bc+ca)+3(x2+y2+z2)=0(a + b + c)^2 - 3(ab + bc + ca) + 3(x^2 + y^2 + z^2) = 0, then which one of the following is correct ?

Read what the stem already gives

If the stem says aa, bb, cc are distinct, then a2+b2+c2−ab−bc−caa^2 + b^2 + c^2 - ab - bc - ca is already known to be positive, and no statement is needed. Data-sufficiency items set this on purpose.
Drill 5 more on sums of squares that vanish

Concept 2 of 3: Least values: t + 1/t is at least 2

For positive tt, t+1t−2=(t−1t)2≥0t + \dfrac1t - 2 = \left(\sqrt t - \dfrac{1}{\sqrt t}\right)^2 \ge 0. So t+1tt + \dfrac1t is smallest, equal to 22, at t=1t = 1. Split a fraction into pieces of that shape and the least value is read off.

Definition

  • For t>0t > 0: t+1t≥2t + \dfrac1t \ge 2, with equality at t=1t = 1.
  • More generally pt+qt≥2pqpt + \dfrac qt \ge 2\sqrt{pq} for p,q,t>0p, q, t > 0 (AM–GM: x+y≥2xyx + y \ge 2\sqrt{xy}).
  • Split: a2+ka+1a=a+k+1a≥k+2\dfrac{a^2 + ka + 1}{a} = a + k + \dfrac1a \ge k + 2.
  • A product of brackets in different positive variables is least when each bracket is least.

AM–GM

pt+qt≥2pq(p,q,t>0)pt + \dfrac{q}{t} \ge 2\sqrt{pq} \quad (p, q, t > 0)

Worked example

Find the least value of x2+4x+4x\dfrac{x^2 + 4x + 4}{x} for x>0x > 0.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2024 · CDS (I) 2024 — Elementary Mathematics · Q60Moderate

Example 2 · Algebraic Identities and Simplification · Sums of Squares and Least Values

What is the minimum value of (a2+3a+1a)(b2+3b+1b)\left(\frac{a^2 + 3a + 1}{a}\right)\left(\frac{b^2 + 3b + 1}{b}\right) for a,b>0a, b > 0 ?

Only for positive values

For negative tt, t+1t≤−2t + \dfrac1t \le -2, so the expression has no least value on all reals. The condition x>0x > 0 in the stem is what makes the answer exist.
Drill 2 more on least values: t + 1/t is at least 2

Concept 3 of 3: Which is larger?

To compare two expressions, look at the sign of their difference. If the difference factors into squares, it is never negative, and it is zero exactly where the squares vanish.

Definition

  • Subtract and factor: if the difference is a square times something positive, the first is never smaller.
  • 'Always greater' fails if the difference can be zero anywhere the question allows.
  • For powers against exponentials (2n2^n and n2n^2), test the small cases one by one; the exponential wins from some point on.

Mean of squares against square of mean

a2+b22−(a+b2)2=(a−b)24≥0\dfrac{a^2 + b^2}{2} - \left(\dfrac{a + b}{2}\right)^2 = \dfrac{(a - b)^2}{4} \ge 0

Worked example

Which is larger, a2+b22\dfrac{a^2 + b^2}{2} or (a+b2)2\left(\dfrac{a + b}{2}\right)^2?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (II) 2019 — Elementary Mathematics · Q29Moderate

Example 3 · Algebraic Identities and Simplification · Sums of Squares and Least Values

The inequality 3N>N33^N > N^3 holds when

Greater or greater-or-equal?

A difference like x2y2(x2−y2)2x^2y^2(x^2 - y^2)^2 is never negative but is zero when x=−yx = -y. If the question asks 'always greater', that single case answers no.
Drill 2 more on which is larger?

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Sums of squares that vanish

    Sum of squared differences

    a2+b2+c2−ab−bc−ca=12[(a−b)2+(b−c)2+(c−a)2]a^2 + b^2 + c^2 - ab - bc - ca = \tfrac12\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right]
  • Least values: t + 1/t is at least 2

    AM–GM

    pt+qt≥2pq(p,q,t>0)pt + \dfrac{q}{t} \ge 2\sqrt{pq} \quad (p, q, t > 0)
  • Which is larger?

    Mean of squares against square of mean

    a2+b22−(a+b2)2=(a−b)24≥0\dfrac{a^2 + b^2}{2} - \left(\dfrac{a + b}{2}\right)^2 = \dfrac{(a - b)^2}{4} \ge 0

Watch out for (3)

Test yourself on Algebraic Identities and Simplification

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