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If $a,b,c$ are positive real numbers and $a+b+c=1$,Prove: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$$ Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.

Things I have done so far:

Using Cauchy inequalities I can write:$$(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})(b+c+a) \geq (a+b+c)^2$$

$a+b+c=1$,So:$$(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})\geq 1$$

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    $\begingroup$ That's not a good direction. $3(a^2+b^2+c^2)\geq (a+b+c)^2=1$. $\endgroup$ – Quang Hoang Jul 31 '14 at 8:21
  • $\begingroup$ @QuangHoang,you are right. the question is wrong.is this the correct form? $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \leq 3(a^2+b^2+c^2)$ $\endgroup$ – user2838619 Jul 31 '14 at 8:25
  • $\begingroup$ I think it's correct. It's just that you overestimated the LHS. $\endgroup$ – Quang Hoang Jul 31 '14 at 8:26
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In this case you want to have quadratic terms in the RHS, so use Cauchy in the following form: $$\left(\sum_{cyc} \frac{a^2}b \right) \left(\sum_{cyc} a^2b \right) \ge \left(a^2+b^2+c^2\right)^2 \tag{1}$$

Then it is enough to show that $$a^2+b^2+c^2 \ge 3\sum_{cyc} a^2b \tag{2}$$ Homogenising, $$\iff \left( a^2+b^2+c^2\right) (a+b+c) \ge 3\sum_{cyc} a^2b $$ $$\iff \sum_{cyc} a^3+ \sum_{cyc} ab^2 \ge 2\sum_{cyc} a^2b $$

which follows from AM-GM as $a^3+ab^2 \ge 2a^2b$.

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lemma: $$\sum_{cyc}\dfrac{a^2}{b}\ge\dfrac{(a+b+c)(a^2+b^2+c^2)}{ab+bc+ac}$$ Proof: $$\Longleftrightarrow (ab+bc+ac)(\sum_{cyc}\dfrac{a^2}{b})\ge (a+b+c)(a^2+b^2+c^2)$$ $$\Longleftrightarrow a^3+b^3+c^3+a^2c+c^2b+b^2a+\sum_{cyc}\dfrac{a^3c}{b}\ge a^3+b^3+c^3+\sum_{sym }a^2b$$ $$\Longleftrightarrow\dfrac{a^3c}{b}+\dfrac{b^3a}{c}+\dfrac{c^3b}{a}\ge ac^2+cb^2+ba^2$$ By AM-GM,we have $$\dfrac{a^3c}{b}+\dfrac{b^3a}{c}\ge 2a^2b$$ $$\dfrac{b^3a}{c}+\dfrac{c^3b}{a}\ge 2b^2c$$ $$\dfrac{a^3c}{b}+\dfrac{c^3b}{a}\ge 2c^2a$$ if we prove this $$\dfrac{a+b+c}{ab+bc+ac}=\dfrac{(a+b+c)^2}{ab+bc+ac}\ge 3$$

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Homogenization gives $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge \frac{3(a^2+b^2+c^2)}{a+b+c}$$ We'll prove a stronger one $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge 6\cdot \frac{a^2+b^2+c^2}{a+b+c}-(a+b+c)$$ $$\Leftrightarrow (a+b+c)\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge 6(a^2+b^2+c^2)-(a+b+c)^2$$ $$\Leftrightarrow (a-b)^2\cdot \frac{a-b+c}{b}+(b-c)^2\cdot \frac{a+b-c}{c}+(c-a)^2\cdot \frac{b+c-a}{a}\ge 0$$ $$\Leftrightarrow S=S_c(a-b)^2+S_a(b-c)^2+S_b(c-a)^2\ge 0$$Since $S_a+S_b=\frac{(a-c)^2+b(a+c)}{ac}\ge 0$

and $S_aS_b+S_bS_c+S_cS_a=\frac{[\sum_{cyc} a^3+3abc-\sum_{cyc} a^2(b+c)]+3abc}{abc}>0$ $$\implies S=\frac{(aS_b+bS_a-cS_a-cS_b)^2+(a-b)^2(S_aS_b+S_bS_c+S_cS_a)}{S_a+S_b}\ge 0$$

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