Inequalities

Take logarithms: the claim is $\ln\left(\sum w_ix_i\right)\ge\sum w_i\ln x_i$, which is Jensen's inequality for the concave function $\ln$ — a chord of a concave graph lies below the graph, so the logarithm of an average beats the average of the logarithms. Equality in Jensen for a strictly concave function forces all the $x_i$ to coincide. With rational weights $w_i=k_i/N$ the statement is ordinary AM–GM on $N$ numbers in which $x_i$ appears $k_i$ times, and that is how it is used in practice: "write $y$ as two copies of $\tfrac y2$". The bookkeeping rule is the equality condition read backwards. AM–GM can only be sharp if all the averaged pieces can be equal at once, so guess the extremal point first, then cut the terms so the pieces agree there and multiply to a constant.

Step 1 — the two-term AM–GM, $x^2+\tfrac{16}{x}\ge2\sqrt{16x}$, is useless: the bound still depends on $x$. A constant bound needs a product of pieces that is constant, so the $x^2$ must be cancelled by two factors of $\tfrac1x$.

Step 2 — split $\tfrac{16}{x}=\tfrac8x+\tfrac8x$:

Step 3 — equality needs the three pieces equal, $x^2=\tfrac8x$, so $x=2$; and indeed $4+8=12$. The minimum is $12$.

How to find it. The exponent $2$ on top and $1$ below say how many copies to make: $x^2$ balances two copies of $x^{-1}$. Calculus agrees — $2x-\tfrac{16}{x^2}$ vanishes at $x=2$ — but AM–GM proves the global bound in one line, with no second-derivative or endpoint check.

Step 1 — the product has weights $1,2,3$, so write the sum as six pieces: $x$, two copies of $\tfrac y2$, three copies of $\tfrac z3$. They still add up to $x+y+z=6$.

Step 2 — AM–GM on the six pieces:

so $xy^2z^3\le4\cdot27=108$.

Step 3 — equality needs $x=\tfrac y2=\tfrac z3$; with $x+y+z=6$ that is $(x,y,z)=(1,2,3)$, and $1\cdot4\cdot27=108$. The maximum is $108$.

How to find it. The exponents are the weights. The maximum of $x^ay^bz^c$ on the plane $x+y+z=s$ splits $s$ in the ratio $a:b:c$ — Lagrange multipliers find that point, and the weighted AM–GM split proves it is the maximum.

Step 1 — decide what each factor should be compared with. The constraint only controls the product of the $a_k$, so the bound for $(1+a_k)^k$ should be a constant times $a_k$ to the first power; then the product of the bounds is a constant.

Step 2 — AM–GM on $k$ pieces, with the $1$ cut into $k-1$ equal parts:

so $(1+a_k)^k\ge\dfrac{k^k}{(k-1)^{k-1}}\,a_k$.

Step 3 — multiply over $k=2,\dots,n$. The constants telescope,

and the $a_k$ multiply to $1$, so the left side is at least $n^n$.

Step 4 — strictness. Equality in Step 2 would need $a_k=\tfrac1{k-1}$ for every $k$, but then $a_2a_3\cdots a_n=\tfrac1{(n-1)!}<1$ when $n\ge3$, which the constraint forbids. So some factor is strict, and the inequality is strict.

How to find it. The exponent $k$ on $1+a_k$ is the number of pieces AM–GM should use, and the constraint says each bound must contain $a_k$ exactly once. Those two facts force the split of $1$ into $k-1$ equal parts. The equality case $a_k=\tfrac1{k-1}$ then explains the strict sign: the only point where every step is tight violates the constraint.

Each step is true, but the bound $4$ is never reached: $a+\tfrac1a=2$ only at $a=1$, and then $b=0$ is not allowed. A bound whose equality case violates the constraint is a true but weak inequality, not the minimum. Expanding instead, the product is $ab+\tfrac1{ab}+\tfrac{a^2+b^2}{ab}$, and with $t=ab$ (so $a^2+b^2=1-2t$) it equals $t+\tfrac2t-2$. Since $t\le\tfrac14$ by AM–GM and $t+\tfrac2t$ decreases on $\left(0,\tfrac14\right]$, the minimum is at $t=\tfrac14$: the value $\tfrac{25}{4}$, at $a=b=\tfrac12$. In a chain of inequalities, write down each step's equality point and check that one admissible point makes them all tight.

The quadratic $q(t)=\sum(a_it-b_i)^2$ is a sum of squares, so it is never negative. Written out it is $At^2-2Bt+C$ with $A=\sum a_i^2$, $B=\sum a_ib_i$, $C=\sum b_i^2$, and a quadratic that never goes negative has discriminant $4B^2-4AC\le0$ — that is Cauchy–Schwarz. Equality means $q$ has a real root $t$, so $b_i=ta_i$ for every $i$: proportionality. Titu is Cauchy–Schwarz applied to $\tfrac{x_i}{\sqrt{y_i}}$ and $\sqrt{y_i}$. Hölder follows from AM–GM: with $A=\sum a_i$ and so on, $\sqrt[3]{\tfrac{a_i}{A}\cdot\tfrac{b_i}{B}\cdot\tfrac{c_i}{C}}\le\tfrac13\left(\tfrac{a_i}{A}+\tfrac{b_i}{B}+\tfrac{c_i}{C}\right)$, and summing over $i$ gives $1$ on the right. The skill is choosing the pairing so that the equality case — proportional vectors — happens where you expect the extremum.

Step 1 — pair $(x,y,z)$ with the coefficient vector $(2,3,6)$:

Step 2 — the left side is $14^2=196$, so $x^2+y^2+z^2\ge4$.

Step 3 — equality needs $(x,y,z)=t(2,3,6)$; substituting, $49t=14$, $t=\tfrac27$, so the minimum $4$ is attained at $\left(\tfrac47,\tfrac67,\tfrac{12}7\right)$. Geometrically $x^2+y^2+z^2$ is the squared distance to the origin, $(2,3,6)$ is the normal of the plane, and the distance from the origin to the plane is $\tfrac{14}{7}=2$.

How to find it. A linear constraint and a sum of squares: the constraint's coefficients are the other vector. The equality case, a point along the normal, is the foot of the perpendicular.

Step 1 — the variables sit in the denominators, so substitute $x=\tfrac1a$, $y=\tfrac1b$, $z=\tfrac1c$; then $xyz=1$ and

using $xyz=1$ in the last step.

Step 2 — each term is now a square over a positive quantity, the shape of Titu:

Step 3 — by AM–GM, $x+y+z\ge3\sqrt[3]{xyz}=3$, so the sum is at least $\tfrac32$.

Step 4 — both steps are tight at $x=y=z=1$, i.e. $a=b=c=1$, where each term is $\tfrac12$.

How to find it. A cube in the denominator together with $abc=1$ invites the reciprocal substitution. Once the numerators are squares and the denominators are linear, Titu is the tool, and the leftover $x+y+z\ge3$ uses the constraint exactly once.

Step 1 — the left side is a product of three one-variable factors, the right side the cube of a sum: that is the shape of Hölder, which wants each factor to be a sum of three terms with one cube among them.

Step 2 — compare each factor with $a^3+2$:

because $a^2-1$ and $a^3-1$ have the same sign for $a>0$ (both negative below $1$, both positive above). So each factor is at least $a^3+1+1$.

Step 3 — put the cubes in different positions and apply Hölder:

Step 4 — equality throughout needs $a=b=c=1$, where both sides equal $27$.

How to find it. Test $a=b=c=1$: both sides are $27$, so every lemma must be tight at $1$. The factor $a^5-a^2+3$ is not a natural Hölder term but $a^3+2$ is, and their difference factors because both $a^5-a^3$ and $a^2-1$ vanish at $a=1$.

Both inequalities are true, but the first is tight only when $a=2b$ and the second only when $a=b$. No admissible point makes both tight, so $4\sqrt2\approx5.66$ is never attained. Pair the terms with the constraint instead: by Titu,

with equality when $\tfrac1a=\tfrac{\sqrt2}{2b}$, i.e. $a=\sqrt2-1$, $b=\tfrac{2-\sqrt2}{2}$ — an admissible point, so $3+2\sqrt2$ is the minimum.