Kvant Math Problem 762
The two inequalities are
Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m21s
Source on kvant.digital
Problem
Prove that for any positive numbers $a$, $b$, $c$, the inequalities $$a+b+c\le\frac{a^2+b^2}{2c}+\frac{b^2+c^2}{2a}+\frac{c^2+a^2}{2b}\le\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}.$$ hold.
Exploration
The two inequalities are
$$S:=\frac{a^2+b^2}{2c}+\frac{b^2+c^2}{2a}+\frac{c^2+a^2}{2b},$$
and
$$a+b+c\le S\le \frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}.$$
First test the symmetric case $a=b=c=t$. Then
$$S=3t,$$
and
$$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}=3t.$$
Hence equality holds throughout when $a=b=c$.
For the left inequality, each summand has the form
$$\frac{x^2+y^2}{2z}.$$
Since $x^2+y^2\ge 2xy$,
$$\frac{x^2+y^2}{2z}\ge \frac{xy}{z}.$$
Summing gives
$$S\ge \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}.$$
The classical inequality
$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c$$
follows from AM-GM, since
$$\frac{ab}{c}+\frac{bc}{a}\ge 2b,$$
and cyclically.
For the right inequality, compare term by term. It is natural to rewrite
$$\frac{a^3}{bc}+\frac{b^3}{ca}$$
against
$$\frac{a^2+b^2}{c}.$$
Indeed,
$$\frac{a^3}{bc}+\frac{b^3}{ca}-\frac{a^2+b^2}{c} =\frac{(a-b)^2(a+b)}{ab,c}\ge 0.$$
This identity is the crucial point. Summing its cyclic analogues should give exactly twice the desired inequality, because every term $(a^2+b^2)/(2c)$ appears naturally.
Problem Understanding
We must prove a double inequality relating three symmetric expressions in positive variables $a,b,c$:
$$a+b+c\le \frac{a^2+b^2}{2c}+\frac{b^2+c^2}{2a}+\frac{c^2+a^2}{2b} \le \frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}.$$
This is a Type B problem, a pure proof.
The core difficulty is the upper bound. The lower bound follows quickly from elementary inequalities, while the upper bound requires finding the correct comparison between the cubic-over-product terms and the quadratic-over-variable terms.
Proof Architecture
Lemma 1. For positive $a,b,c$,
$$\frac{a^2+b^2}{2c}\ge \frac{ab}{c}.$$
This is the standard inequality $a^2+b^2\ge 2ab$.
Lemma 2. For positive $a,b,c$,
$$\frac{ab}{c}+\frac{bc}{a}\ge 2b,$$
and its cyclic analogues.
This is AM-GM applied to the two positive terms.
Lemma 3. For positive $a,b,c$,
$$\frac{a^3}{bc}+\frac{b^3}{ca}\ge \frac{a^2+b^2}{c}.$$
After multiplying by $abc$, the difference becomes
$(a-b)^2(a+b)\ge0$.
The lower bound follows from Lemmas 1 and 2. The upper bound follows by summing the three cyclic instances of Lemma 3 and dividing by $2$.
The most delicate point is Lemma 3, because the proof depends on an exact algebraic factorization.
Solution
Let
$$S=\frac{a^2+b^2}{2c}+\frac{b^2+c^2}{2a}+\frac{c^2+a^2}{2b}.$$
We first prove that $S\ge a+b+c$.
Since $a^2+b^2\ge 2ab$,
$$\frac{a^2+b^2}{2c}\ge \frac{ab}{c}.$$
Applying the same inequality cyclically,
$$S\ge \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}.$$
Now
$$\frac{ab}{c}+\frac{bc}{a}\ge 2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}} =2b.$$
Similarly,
$$\frac{bc}{a}+\frac{ca}{b}\ge 2c,$$
and
$$\frac{ca}{b}+\frac{ab}{c}\ge 2a.$$
Adding these three inequalities yields
$$2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right) \ge 2(a+b+c),$$
hence
$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \ge a+b+c.$$
Combining this with the previous estimate gives
$$S\ge a+b+c.$$
Next we prove the upper bound.
Consider
$$\frac{a^3}{bc}+\frac{b^3}{ca}-\frac{a^2+b^2}{c}.$$
Multiplying by $abc$, we obtain
$$a^4+b^4-a^3b-ab^3.$$
Factoring,
$$a^4+b^4-a^3b-ab^3 =(a-b)^2(a^2+ab+b^2)-ab(a-b)^2 =(a-b)^2(a^2+b^2) =(a-b)^2(a+b)^2-2ab(a-b)^2.$$
A simpler factorization is
$$a^4+b^4-a^3b-ab^3 =(a-b)^2(a^2+ab+b^2+ab) =(a-b)^2(a+b)^2.$$
Hence
$$a^4+b^4-a^3b-ab^3\ge0,$$
and therefore
$$\frac{a^3}{bc}+\frac{b^3}{ca} \ge \frac{a^2+b^2}{c}.$$
Applying the same argument cyclically,
$$\frac{b^3}{ca}+\frac{c^3}{ab} \ge \frac{b^2+c^2}{a},$$
and
$$\frac{c^3}{ab}+\frac{a^3}{bc} \ge \frac{c^2+a^2}{b}.$$
Adding these three inequalities gives
$$2\left( \frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab} \right) \ge \frac{a^2+b^2}{c} +\frac{b^2+c^2}{a} +\frac{c^2+a^2}{b}.$$
Dividing by $2$,
$$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab} \ge \frac{a^2+b^2}{2c} +\frac{b^2+c^2}{2a} +\frac{c^2+a^2}{2b} =S.$$
Thus
$$a+b+c\le S\le \frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the passage from
$$S\ge \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}$$
to
$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c.$$
A careless application of AM-GM to the three terms would only give
$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge 3,$$
which is dimensionally inappropriate. The correct argument pairs the terms:
$$\frac{ab}{c}+\frac{bc}{a}\ge 2b,$$
and cyclically, after which summation yields the desired inequality.
The second delicate step is the factorization in Lemma 3. Starting from
$$a^4+b^4-a^3b-ab^3,$$
one must check carefully that
$$a^4+b^4-a^3b-ab^3 =(a-b)^2(a+b)^2.$$
Expanding the right-hand side gives
$$(a^2-2ab+b^2)(a^2+2ab+b^2) =(a^2+b^2)^2-(2ab)^2 =a^4+b^4-2a^2b^2,$$
which is not the same expression. Thus that factorization is incorrect.
The correct computation is
$$a^4+b^4-a^3b-ab^3 =(a-b)(a^3-b^3) =(a-b)^2(a^2+ab+b^2)\ge0.$$
This is the identity actually needed for the proof.
The third delicate step is the summation of the cyclic inequalities. Each term $\frac{a^3}{bc}$ appears in exactly two of them, so the left-hand side becomes
$$2\left(\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\right),$$
which explains the final division by $2$.
Alternative Approaches
The lower bound can also be derived directly from Cauchy-Schwarz in Engel form:
$$\frac{a^2+b^2}{2c} = \frac{a^2}{2c}+\frac{b^2}{2c},$$
so that
$$S=\frac12\sum_{\text{cyc}} \left(\frac{a^2}{b}+\frac{a^2}{c}\right).$$
Applying Titu's lemma to the six terms gives
$$S\ge \frac{(2a+2b+2c)^2}{2(a+b+c+a+b+c)} =a+b+c.$$
For the upper bound, one may write
$$\frac{a^3}{bc}-\frac{a^2}{2b}-\frac{a^2}{2c} = \frac{a^2\bigl(2a-b-c\bigr)}{2bc}$$
and sum cyclically. The resulting expression can then be transformed into
$$\frac12\sum_{\text{cyc}} \frac{(a-b)^2(a+b)}{ab},$$
which is nonnegative. The main proof is preferable because it uses a simple two-variable inequality whose cyclic sum immediately produces the required estimate.