Let $a$, $b$, $c$ be three real positive numbersand $a^2 + b^2 + c^2 =3$. Find the minimum of the expression $$P = \dfrac{a^2}{b + 2c} +\dfrac{b^2}{c + 2a}+ \dfrac{c^2}{a + 2b}.$$ I tried $$\dfrac{a^2}{b + 2c} +\dfrac{b^2}{c + 2a}+ \dfrac{c^2}{a + 2b} \geqslant \dfrac{(a + b+c)^2}{3(a + b + c)}$$ or $$P \geqslant \dfrac{a + b + c}{3}$$
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Can you provide more steps on how you have arrived at $P \geqslant \dfrac{a+b+c}{3}$ ? $\endgroup$– lspJun 10, 2013 at 11:05
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
$P\geq \frac{9}{a^2(b+2c)+b^2(c+2a)+c^2(a+2b)}$ by Cauchy-Schwarz inequality. $a^2b+b^2c+ac^2\leq\sqrt{3(a^2b^2+a^2c^2+b^2c^2)}$ and $2(a^2c+ab^2+bc^2)\leq 2\sqrt{3(a^2b^2+a^2c^2+b^2c^2)}$, again by Cauchy-Schwarz inequality and $P\geq \frac{\sqrt{3}}{\sqrt{(a^2b^2+a^2c^2+b^2c^2)}}$. Arithmetic-geometric inequality gives $(a^2b^2+a^2c^2+b^2c^2)\leq \frac{(a^2+b^2+c^2)^2}{3}=3$, so finally $P\geq 1$.
-
$\begingroup$ More detailed, $(a^2b+b^2c+ac^2)^2=(a ab+b bc+ c ac)^2\leq (a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)$, which is equal to $3(a^2b^2+b^2c^2+c^2a^2)$, the method is same for other inequality. The last inequality in my answer is equivalent with $3(xy+yz+zx)\leq (x+y+z)^2$, which is easy to show. $\endgroup$– alansJun 10, 2013 at 11:36