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Let $a, b, c, d$ be positive real numbers such that $a^2 + b^2 + c^2 + d^2 = 4$, show that $$\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge 4.$$

My try:

$$\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge a + b + c + d,$$

yet $$a + b + c + d \le \sqrt{4(a^2 + b^2 + c^2 + d^2)} = 4.$$

Thus, direct application of Cauchy-Schwarz inequality is too weak. I tried other methods but with no significant progress:

$$(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a)^2 \ge \frac {(a^{4/3} + b^{4/3} + c^{4/3})^3}{a^2 + b^2 + c^2 + d^2} = \frac {(a^{4/3} + b^{4/3} + c^{4/3})^3}4.$$

I also observed that

$$(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a) + (\frac {a^2}c + \frac {b^2}d + \frac {c^2}a + \frac {d^2}b) + (\frac {a^2}d + \frac {b^2}a + \frac {c^2}b + \frac {d^2}c) + (\frac {a^2}a + \frac {b^2}b + \frac {c^2}c + \frac {d^2}d) = 4 (\frac 1a + \frac 1b + \frac 1c + \frac 1d) \ge 16,$$

since $$\frac 1a + \frac 1b + \frac 1c + \frac 1d \ge \sqrt{\frac {(1 + 1 + 1 + 1)^3}{a^2 + b^2 + c^2 + d^2}} = 4.$$

Now my work might seem stupid or off-topic here, but I provide it here because I wish any of these attempts will lead to a solution. Any hints will be appreciated.

Yuxiao Xie
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  • Sharing your effort is much appreciated. CS should work, perhaps not in the way you tried. Will get back to this later today in case you haven't got an answer by then. – Macavity Apr 15 '16 at 14:56
  • @Macavity: It seems to me that your (deleted) answer http://math.stackexchange.com/a/856255/42969 fits here. – Martin R Apr 15 '16 at 15:03
  • Unfortunately I can't see deleted posts on the mobile app @MartinR But do post if that works. – Macavity Apr 15 '16 at 15:06

4 Answers4

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(This is actually from the deleted answer to a different question, posted here with permission.)

From Cauchy-Schwarz: $$ \left(\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \right) \left( a^2 b + b^2 c + c^2 d + c^2 a \right) \ge (a^2+b^2+c^2+d^2)^2 =16 $$ therefore it suffices to show that $$ \tag{*} \left( a^2 b + b^2 c + c^2 d + c^2 a \right) \le 4 $$ Using Cauchy-Schwarz again: $$ \left( a^2 b + b^2 c + c^2 d + c^2 a \right)^2 \le (a^ 2+ b^2 + c^2 + d^2)(a^2b^2 + b^2c^2 + c^ 2d^2 + d^2 a^2) \\ = 4 (a^2+c^2)(b^2+d^2) \\ \le 4 \left( \frac {a^2+b^2+c^2+d^2}{2} \right) ^2 = 16 $$ with AM-GM in the last step. From this $(*)$ follows.

Community
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Martin R
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6

Just another way is to use Holder : $$\left(\sum_{cyc} \frac{a^2}b \right)^2 \left( \sum_{cyc} a^2b^2\right) \geqslant \left( \sum_{cyc} a^2\right)^3=4^3$$

So it remains to show $$a^2b^2+b^2c^2+c^2d^2+d^2a^2 = (a^2+c^2)(b^2+d^2) \leqslant \frac14(a^2+b^2+c^2+d^2)^2 = 4 $$

Macavity
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4

This may also be shown using Jensen's inequality with $x \mapsto \frac{1}{\sqrt{x}}$ and weights $\frac{a^2}{4}$, $\frac{b^2}{4}$, $\frac{c^2}{4}$, $\frac{d^2}{4}$, which by the given assumption sum to one.

\begin{align*} \frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a &= 4 \left( \frac{a^2}{4}\frac{1}{\sqrt{b^2}} + \frac{b^2}{4}\frac{1}{\sqrt{c^2}} + \frac{c^2}{4}\frac{1}{\sqrt{d^2}}+ \frac{d^2}{4}\frac{1}{\sqrt{a^2}} \right) \\ &\geq 4 \left( \frac{1}{\sqrt{ \frac{a^2}{4} b^2 + \frac{b^2}{4} c^2 + \frac{c^2}{4} d^2 + \frac{d^2}{4} a^2} } \right) \end{align*}

So it suffices to show that \begin{equation} \frac{a^2}{4} b^2 + \frac{b^2}{4} c^2 + \frac{c^2}{4} d^2 + \frac{d^2}{4} a^2 \leq 1 \end{equation}

This is the same ending point as the other existing answers, which factor and use AM-GM to finish.

$$\frac{a^2}{4} b^2 + \frac{b^2}{4} c^2 + \frac{c^2}{4} d^2 + \frac{d^2}{4} a^2 = \frac{1}{4} (a^2+c^2)(b^2+d^2) \leq \frac{1}{4^2}(a^2+b^2+c^2+d^2)^2 = 1 $$

A. Webb
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    Nice +1. Once you have a convexity argument possibilities with mean inequalities abound. AM-GM should also do the trick, just add a bunch of $a^2/b+a^2b\ge 2a^2$ together, and finally we are left to show $\sum a^2b\le 4$ again! – Macavity Apr 15 '16 at 19:49
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by the AM-GM inequality:

Given that: 4= aˆ2 + bˆ2 + cˆ2 + dˆ2 ===> 1 = 1/4 (aˆ2 + bˆ2 + cˆ2 + dˆ2) ≥ 4^√((aˆ2)(bˆ2)(cˆ2)(dˆ2))= √(abcd)

since: 1≥ √(abcd)

(aˆ2+bˆ2+cˆ2+dˆ2)/4 (aˆ2/b + bˆ2/c + cˆ2/d + dˆ2/a)≥ 4ˆ√(abcd)(aˆ2 + bˆ2 + cˆ2 + dˆ2)ˆ4 = 4√(abcd)≥ 4

Hence (aˆ2/b + bˆ2/c + cˆ2/d + dˆ2/a) ≥ 4

Hcb
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