## Task 15/65

What restriction does $\mu$ have to satisfy if the inequality is to hold for all real $a, b$ and $\lambda$:

$$
a^{4}+\lambda^{2} b^{4} \geq \mu a^{2} b^{2}
$$
First, it is clear that the inequality holds for $a=0$ or for $b=0$ or for $a=b=0$ for any real $\lambda$. Excluding these cases, for further considerations, one can assume $a \neq 0 ; b \neq 0$, divide by $a^{4}$, and substitute $\frac{b^{2}}{a^{2}}=x$; thus, one obtains

$$
1+\lambda^{2} x^{2} \geq \mu x \quad 1-\mu x+\lambda^{2} x^{2} \geq 0
$$

From this, one can easily return to the given inequality. The task now is to determine under what conditions the quadratic form

$$
y=1-\mu x+\lambda^{2} x^{2}
$$

is non-negative. One completes the first terms to a square and finds

$$
y=\left(1-\frac{\mu}{2} x\right)^{2}+\left(\lambda^{2}-\frac{\mu^{2}}{4}\right) x^{2}
$$

This sum of squares of real numbers is non-negative if the coefficients of the squares are, i.e., if

$$
\lambda^{2}-\frac{\mu^{2}}{4} \geq 0 ; \quad \mu \leq 2|\lambda|
$$

holds. This is the desired condition. However, equality in (3) does not imply equality in (1), as in (2b) the first square remains and one has

$$
\left(1-\frac{\mu}{2} x\right)^{2} \geq(1-|\lambda| x)^{2} \geq 0
$$

and equality in (1) only occurs for

$$
x=\frac{1}{|\lambda|}
$$

If (3) is not satisfied, i.e., $\mu>2|\lambda|$, the inequality does not hold for all $a$ and $b$; a neighborhood of the critical value (4) must be excluded. This is easily recognized if (2) is interpreted as a parabola whose axis runs in the direction of the positive y-axis of a right-angled Cartesian coordinate system. The points of intersection of the parabola with the x-axis are at

$$
x_{1 ; 2}=\frac{1}{2 \lambda^{2}}\left(\mu \pm \sqrt{\mu^{2}-4 \lambda^{2}}\right)
$$

For $\mu>2|\lambda|$, $y$ is non-negative only if $x$ does not lie between $x_{1}$ and $x_{2}$ according to (5).

Summary:

If $a=0$ or $b=0$ or $a=b=0$, the inequality holds unconditionally for any $\lambda$ and $\mu$. If $a \neq 0$ and $b \neq 0$ and

1. $\mu \leq 2|\lambda|$, then it holds for all $a, b$ and $\lambda$.
2. $\mu=2|\lambda|$ or $|\lambda|=x=\frac{b^{2}}{a^{2}}$, then equality holds.
3. $\mu>2|\lambda|$, then the inequality holds only for those $a$ and $b$ for which $x=\frac{b^{2}}{a^{2}}$ does not lie in the open interval $x_{1}<x<x_{2}$.