Kvant Math Problem 1193
Let
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m10s
Source on kvant.digital
Problem
Prove (for any numbers $a$, $b$, $c$, $x$, $y$, $z$) the inequality $$\begin{gather*} ax+by+cz+\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}\ge\ \ge\frac23(a+b+c)(x+y+z). \end{gather*}$$
V. Kyrtojа (Romania)
Exploration
Let
$$u=(a,b,c),\qquad v=(x,y,z).$$
Then the inequality becomes
$$u\cdot v+|u|,|v|\ge \frac23 (a+b+c)(x+y+z).$$
The left-hand side resembles the Cauchy-Schwarz inequality. Since
$$(a+b+c)(x+y+z) =(u\cdot e)(v\cdot e),$$
where $e=(1,1,1)$, the right-hand side is expressed through projections onto the vector $e$.
A first thought is to use
$$(u\cdot e)^2\le |u|^2|e|^2=3|u|^2,$$
and similarly for $v$. This gives
$$|(a+b+c)(x+y+z)| \le 3|u|,|v|.$$
Hence
$$\frac23 |(a+b+c)(x+y+z)| \le 2|u|,|v|.$$
To obtain the desired inequality it would then suffice to prove
$$u\cdot v+|u|,|v| \ge \frac23 |(a+b+c)(x+y+z)|.$$
Using the previous estimate, this reduces to
$$u\cdot v+|u|,|v|\ge 0,$$
which is exactly the Cauchy-Schwarz inequality in the form
$$u\cdot v\ge -|u|,|v|.$$
The potentially dangerous step is replacing the right-hand side by its absolute value. Since the desired inequality has no absolute value, we must check carefully that
$$\frac23(a+b+c)(x+y+z) \le \frac23 |(a+b+c)(x+y+z)|$$
always holds. Once this is established, the argument is complete.
Problem Understanding
We must prove that for arbitrary real numbers $a,b,c,x,y,z$,
$$ax+by+cz+\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)} \ge \frac23(a+b+c)(x+y+z).$$
This is a Type B problem, a pure proof.
The core difficulty is relating the expression involving the scalar product and Euclidean norms on the left to the product of sums on the right. The natural tools are the Cauchy-Schwarz inequality applied both to the vectors $(a,b,c)$ and $(x,y,z)$ and to the vector $(1,1,1)$.
Proof Architecture
Lemma 1. For vectors $u=(a,b,c)$ and $v=(x,y,z)$,
$$u\cdot v+|u|,|v|\ge 0.$$
This follows from the Cauchy-Schwarz inequality $u\cdot v\ge -|u|,|v|$.
Lemma 2. The sums satisfy
$$|(a+b+c)(x+y+z)|\le 3|u|,|v|.$$
Apply Cauchy-Schwarz to $u$ and $(1,1,1)$, and separately to $v$ and $(1,1,1)$.
Lemma 3. Consequently,
$$\frac23(a+b+c)(x+y+z)\le 2|u|,|v|.$$
This is obtained from Lemma 2 and the inequality $t\le |t|$.
The hardest point is ensuring that the passage from $(a+b+c)(x+y+z)$ to its absolute value preserves the required inequality.
Solution
Set
$$u=(a,b,c),\qquad v=(x,y,z).$$
Then
$$u\cdot v=ax+by+cz,$$
and
$$|u|=\sqrt{a^2+b^2+c^2},\qquad |v|=\sqrt{x^2+y^2+z^2}.$$
By the Cauchy-Schwarz inequality,
$$u\cdot v\ge -|u|,|v|,$$
hence
$$u\cdot v+|u|,|v|\ge 0.$$
Therefore
\begin{equation}
u\cdot v+|u|,|v|+2|u|,|v|
\ge 2|u|,|v|.
\tag{1}
\end{equation}
Now let $e=(1,1,1)$. Since $|e|=\sqrt3$, another application of Cauchy-Schwarz gives
$$|u\cdot e| \le |u|,|e| =\sqrt3,|u|,$$
that is,
$$|a+b+c|\le \sqrt3,|u|.$$
Similarly,
$$|x+y+z|\le \sqrt3,|v|.$$
Multiplying these inequalities, we obtain
$$|(a+b+c)(x+y+z)| \le 3|u|,|v|.$$
Hence
$$\frac23(a+b+c)(x+y+z) \le \frac23|(a+b+c)(x+y+z)| \le 2|u|,|v|.$$
Combining this with (1), we get
$$u\cdot v+|u|,|v|+2|u|,|v| \ge \frac23(a+b+c)(x+y+z).$$
Subtracting $2|u|,|v|$ from the left-hand side yields
$$u\cdot v+|u|,|v| \ge \frac23(a+b+c)(x+y+z).$$
Returning to the original notation,
$$ax+by+cz+\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)} \ge \frac23(a+b+c)(x+y+z).$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is
$$u\cdot v+|u|,|v|\ge 0.$$
Starting from Cauchy-Schwarz,
$$|u\cdot v|\le |u|,|v|,$$
so
$$-|u|,|v|\le u\cdot v.$$
Adding $|u|,|v|$ to both sides gives the claim. Equality occurs when $v$ is a negative scalar multiple of $u$.
The second delicate step is
$$|(a+b+c)(x+y+z)|\le 3|u|,|v|.$$
Applying Cauchy-Schwarz separately,
$$|a+b+c| =|u\cdot e| \le |u|\sqrt3,$$
and
$$|x+y+z| =|v\cdot e| \le |v|\sqrt3.$$
Multiplication yields the factor $3$. Forgetting one factor $\sqrt3$ would produce the incorrect constant.
The third delicate step is replacing the product by its absolute value. Since every real number $t$ satisfies $t\le |t|$,
$$\frac23(a+b+c)(x+y+z) \le \frac23|(a+b+c)(x+y+z)|.$$
Without this observation, the argument would not connect the right-hand side with the norm estimate obtained from Cauchy-Schwarz.
Alternative Approaches
Introduce the unit vector
$$n=\frac1{\sqrt3}(1,1,1).$$
Then
$$\frac23(a+b+c)(x+y+z) =2(u\cdot n)(v\cdot n).$$
The desired inequality becomes
$$u\cdot v+|u|,|v| \ge 2(u\cdot n)(v\cdot n).$$
Decompose
$$u=\alpha n+p,\qquad v=\beta n+q,$$
where $p\perp n$ and $q\perp n$. The inequality reduces after substitution to
$$\alpha\beta+\sqrt{(\alpha^2+|p|^2)(\beta^2+|q|^2)} \ge 2\alpha\beta.$$
Thus it remains to show
$$\sqrt{(\alpha^2+|p|^2)(\beta^2+|q|^2)} \ge \alpha\beta.$$
This follows directly from the nonnegativity of
$$(\alpha^2+|p|^2)(\beta^2+|q|^2)-\alpha^2\beta^2.$$
The main proof is preferable because it uses only two standard applications of Cauchy-Schwarz and reaches the result with minimal computation.