Olympiad Inequalities: Tips from Engel’s book

In this post we focus on inequalities. First we look at tips from Engel’s book, then we solve a couple of problems using these tips. Finally I leave you with an exercise set.

Tips from Engel’s book

Strategies for Proving Inequalities:

1. Try to transform the inequality into the form
   \[
   \sum p_i, \quad p_i > 0,
   \]
   e.g., \( P_i = x_i^2 \).
2. Does the expression remind you of the AM, GM, HM, or QM?
3. Can you apply the Cauchy–Schwarz inequality? This is especially tricky.
   You can apply this inequality far more often than you think.
4. Can you apply the Rearrangement inequality? Again, this theorem is much underused.
   You can apply it in most unexpected circumstances.
5. Is the inequality symmetric in its variables \( a, b, c, \ldots \)?
   In that case, assume \( a \leq b \leq c \leq \ldots \). Sometimes one can assume that \( a \) is the maximal or minimal element.
   It may be advantageous to express the inequality using elementary symmetric functions.
6. An inequality homogeneous in its variables can be normalized.
7. If you are dealing with an inequality for the sides \( a, b, c \) of a triangle, think of the triangle inequality in its many forms.
   Especially, try the substitution:
   \[
   a = x + y, \quad b = y + z, \quad c = z + x \quad \text{with } x, y, z > 0.
   \]
8. Bring the inequality into the form
   \[
   f(a, b, c, \ldots) > 0.
   \]
   Is \( f \) quadratic in one of its variables? Can you find its discriminant?

Problems

Prove that for any triangle with sides $a,b,c$ and area $A$, $$a^2 + b^2 + c^2 \geq 4\sqrt{3}A.$$

Proof: We will write down the steps and explain it below: \[a^2 + b^2 +c^2 \geq ab+bc+ca = \dfrac{2A}{\sin C}+\dfrac{2A}{\sin A}+\dfrac{2A}{\sin B} \geq \dfrac{18A}{\sin A + \sin B + \sin C} \geq \dfrac{36A}{3\sqrt{3}} = 4\sqrt{3}A.\]
Equality iff $a=b=c$.

The explanations for each step

1. First inequality follows from applying AM-GM to the sets $\{a^2,b^2\},\{b^2,c^2\},\{c^2,a^2\}$ and adding the inequalities.
2. The first equality follows from the area formula in terms of sines.
3. The second inequality follows from AM-HM.
4. The third inequality follows from Jensen’s inequality applied to sin A, sin B and sin C noting that sin is concave in $[0,\pi]$ and $A+B+C=\pi$.

Remark: There are five other proofs of this fact in Engel’s problem solving strategies.

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In triangle ABC, the angle bisectors $AD, BE, CF$ meet at $I$. Show that \[\dfrac14 < \dfrac{IA}{AD}\cdot \dfrac{IB}{BE} \cdot \dfrac{IC}{CF} \leq \dfrac{8}{27}.\]

Proof: By angle bisector theorem, we see that \[\dfrac{IA}{ID} = \dfrac{b+c}{a}\] and thus \[\dfrac{AI}{AD} = \dfrac{b+c}{a+b+c}.\] So we have to show \[\dfrac14 < \dfrac{(b+c)(a+b)(c+a)}{(a+b+c)^3}\leq \dfrac{8}{27}.\]
The upper bound follows from applying AM-GM to $a+b,b+c,c+a$.

For the lower bound, define
$x = a+b-c, y=b+c-a, z=c+a-b$ and note that $2a = x+z, 2b=x+y, 2c=y+z.$
By triangle inequality, we note that $x,y,z>0.$ Since the inequality is homogenous, we will normalise and impose the constraint $a+b+c=1$ which implies $x+y+z=1$.

Thus we have
\[ \dfrac{(b+c)(a+b)(c+a)}{(a+b+c)^3} = (1-a)(1-b)(1-c) = \dfrac{(2-x-z)(2-x-y)(2-y-z)}8\]

\[= \dfrac{(1+y)(1+z)(1+x)}{8} =\dfrac{(1+x+y+z+ xy+yz+zx+xyz)}{8} > \dfrac28 = \dfrac14.\]

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Here are some practice problems from Arthur Engel’s book:

Solve the problems in the comments.