Prove that that $\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c} \geq \dfrac{(x+y+z)^2}{a+b+c}.$ with $a,b,c$ positive real numbers.


I tried using Cauchy-Schwarz, but I can't find the correct $a_i$ and $b_i$. How would you solve this using Cauchy-Schwarz?

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    $\begingroup$ with $a,b,c$ positive reals and $x,y,z$ real numbers $\endgroup$ – Dr. Sonnhard Graubner Jan 4 '16 at 19:19
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    $\begingroup$ $$(a+b+c)\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)\ge \left(\sqrt{a}\sqrt{\frac{x^2}{a}}+ \sqrt{b}\sqrt{\frac{y^2}{b}}+\sqrt{c}\sqrt{\frac{z^2}{c}}\right)^2=(x+y+z)^2$$ $\endgroup$ – user236182 Jan 4 '16 at 19:24
  • $\begingroup$ See the proof in math.stackexchange.com/questions/1594286/… $\endgroup$ – Gordon Jan 4 '16 at 19:28
  • $\begingroup$ Titu's lemma (brilliant.org/wiki/titus-lemma) and the Cauchy-Schwarz inequality are essentially equivalent. It is interesting to point out that Titu's lemma for two variables (almost trivial) plus induction give a neat proof of the general Cauchy-Schwarz inequality. $\endgroup$ – Jack D'Aurizio Jan 4 '16 at 19:49

More generally, the following is called Titu's Lemma or Engel's form of Cauchy-Schwarz inequality:

For all $a_i\in\mathbb R$, $b_i\in\mathbb R^+$:

$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\cdots+a_n)^2}{b_1+b_2+\cdots+b_n}$$

Proof: by Cauchy-Schwarz: $$(b_1+b_2+\cdots+b_n)\left(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n}\right)$$

$$\ge \left(\sqrt{b_1}\sqrt{\frac{a_1^2}{b_1}}+\sqrt{b_2}\sqrt{\frac{a_2^2}{b_2}}+\cdots+\sqrt{b_1}\sqrt{\frac{a_n^2}{b_n}}\right)^2$$

$$=(|a_1|+|a_2|+\cdots+|a_n|)^2\ge (a_1+a_2+\cdots+a_n)^2$$

with equality if and only if $\frac{a_1^2}{b_1^2}=\frac{a_2^2}{b_2^2}=\cdots=\frac{a_n^2}{b_n^2}$ and $(|a_1|+|a_2|+\cdots+|a_n|)^2= (a_1+a_2+\cdots+a_n)^2$, i.e. if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots=\frac{a_n}{b_n}$.

  • $\begingroup$ Another proof is by induction in $n$. Equality is satisfied iff $\frac{a_i}{b_i}=\frac{a_j}{b_j}$ for any $i,j$ $\endgroup$ – sinbadh Jan 4 '16 at 19:38
  • $\begingroup$ @sinbadh Should be $\left|\frac{a_i}{b_i}\right|=\left|\frac{a_j}{b_j}\right|$. $\endgroup$ – user236182 Jan 4 '16 at 19:45
  • $\begingroup$ That's correct. But in the last equality, it should be $(|a_1|+...+|a_n|)^2$ $\endgroup$ – sinbadh Jan 4 '16 at 19:51
  • $\begingroup$ @sinbadh right, edited. $\endgroup$ – user236182 Jan 4 '16 at 19:52
  • $\begingroup$ But... if you don't want abs function, the statement $(|a_1|+\cdots+|a_n|)^2=(a_1+\cdots a_n)^2$ is equivalent to all $a_i$ have the same sign, and therefore, equality is satisfied iff $\dfrac{a_i}{b_i}=\dfrac{a_j}{b_j}$ $\endgroup$ – sinbadh Jan 4 '16 at 19:55

HINT: the left-hand side minus the right-hand side is equal to $${z}^{2}{a}^{2}b+{y}^{2}{a}^{2}c+{z}^{2}a{b}^{2}-2\,abcxy-2\,abcxz-2\,a bcyz+{y}^{2}a{c}^{2}+{x}^{2}{b}^{2}c+{x}^{2}b{c}^{2} \geq 0$$ can you proceed? (sum of squares!)

  • $\begingroup$ Sonnhard Grauber I thought you said this could be done by Cauchy-Schwarz. $\endgroup$ – Jacob Willis Jan 4 '16 at 19:25
  • $\begingroup$ @JacobWillis $$(a+b+c)\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)\ge \left(\sqrt{a}\sqrt{\frac{x^2}{a}}+ \sqrt{b}\sqrt{\frac{y^2}{b}}+\sqrt{c}\sqrt{\frac{z^2}{c}}\right)^2=(x+y+z)^2$$ $\endgroup$ – user236182 Jan 4 '16 at 19:26
  • $\begingroup$ yes you see above a proof by Cauchy Schwarz $\endgroup$ – Dr. Sonnhard Graubner Jan 4 '16 at 19:27

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