# maximum value of algebraic expression (another)

if $$p^2+q^2+r^2=5$$ and $$p,q,r$$ all are real number,

then maximum value of $$(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$

what i try . Expanding $$(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$

$$41p^2+41q^2+25r^2-24pq-40qr-30pr$$

$$25\times 5+16p^2+16q^2-24pq-40qr-30pr$$

How i use inequality to find maximum of given expression

• The maximum is $250$, achieved at $p=-\frac{4}{\sqrt{5}},\ q =\frac{3}{\sqrt{5}},\ r=0$. – Federico Dec 17 '18 at 19:40
• you expanded the thing incorrectly. For example, the coefficient of $p^2$ is actually $16+25=41$ – Will Jagy Dec 17 '18 at 20:01
We'll prove that $$250$$ it's a maximal value.
Indeed, $$250\geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ it's $$50(p^2+q^2+r^2)\geq(4p-3q)^2+(5q-4r)^2+(5p-3r)^2$$ or $$(3p+4q+5r)^2\geq0.$$ The equality occurs for $$p^2+q^2+r^2=5$$ and $$3p+4q+5r=0.$$