# Inequality in real numbers.

Let $a,b,c,d$ be real numbers. If $a^2+b^2\leq2$ and $c^2+d^2\leq4$, the maximum value of the expression $ac+bd$ is?

How to proceed with it. I was given hint to use Cauchy–Schwarz inequality, but couldn't solve it.

• Yes but please give a complete solution – sayan Jun 8 '17 at 10:38

Hint. By Cauchy–Schwarz inequality, $$(ac+bd)^2\leq (a^2+b^2)(c^2+d^2).$$ When does the equality holds? Note that $(ac+bd)^2\leq M$ implies that $-\sqrt{M}\leq ac+bd\leq \sqrt{M}$.

• @lulu Thanks a lot! – Robert Z Jun 8 '17 at 10:42
• What is the maximum value? – sayan Jun 8 '17 at 10:48
• @sayan Would you mind to make a try? – Robert Z Jun 8 '17 at 10:50
• It is continually comin $2√2$. I am not getting 3 – sayan Jun 8 '17 at 10:52
• Where does $3$ come from? – Robert Z Jun 8 '17 at 10:53

Hint: By AM-GM,

$$ab+cd\leq\frac{a^2+b^2}{2}+\frac{c^2+d^2}{2}.$$

By Cauchy-Schwarz,

$$(ac+bd)^2\leq(a^2+b^2)(c^2+d^2).$$

• It is $ac+bd$. Sorry for the mistake – sayan Jun 8 '17 at 10:40

By Causchy-Schwartz:

$$2\cdot 4\ge(a^2+b^2)(c^2+d^2)\ge(ac+bd)^2$$

CS says that $(a^2+b^2)(c^2+d^2)\ge(ab+cd)^2$

After using C-S we get a value $2\sqrt2$, which occurs for $(a,b)||(c,d)$, $|(ab)|=\sqrt2$ and $|(c,d)|=2$, which happens for $a=b=1$ and $c=d=\sqrt2$.

Thus, $2\sqrt2$ is the answer.

• What does $(a,b)||(c,d)$ mean – sayan Jun 8 '17 at 17:45
• @sayan It's just vector $(a,b)$ parallel to vector $(c,d)$. – Michael Rozenberg Jun 8 '17 at 18:08