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Let $a,b,c,x,y,z >0$ and $A=a^2+b^2+c^2,\ B=x^2+y^2+z^2,\ C=ax+by+cz$. By Cauchy-Schwarz inequality, we always have $C^2\le AB$. If $C^2<AB$, prove that $$ \frac{A}{C+\sqrt{2(AB-C^2)}}<\frac{a+b+c}{x+y+z}<\frac{C+\sqrt{2(AB-C^2)}}{B}. $$ I created this inequality. Are there any nice proofs?

user1551
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math110
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    Doesn't Cauchy-Schwarz guaranteee that case (2) never happens? – darij grinberg Mar 28 '13 at 01:36
  • yes,The problem from this if $C^2=AB$, then we use Cauchy-Schwarz have $\dfrac{a+b+c}{x+y+z}=\dfrac{C^2}{B^2}$ – math110 Mar 28 '13 at 01:45
  • Are you sure that the inequality you invented is true ? – Ewan Delanoy May 22 '13 at 08:53
  • I have somewhat modified your question. As pointed out by darij grinberg, case (2) never happens and so I omitted it. Also, I redefined $A,B,C$ as the squares of your original $A,B,C$, so as to avoid carrying unnecessary exponents. If you think my edit is inappropriate, please feel free to roll back. – user1551 May 22 '13 at 08:56

1 Answers1

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It seems the following.

The inequality is not strict. For instance, put $a=b=c=x=1$, $y=z=0$. Then $A=3$, $B=C=1$, and $C^2<AB$. But

$$\frac{a+b+c}{x+y+z}=3=\frac{C+\sqrt{2(AB-C^2)}}{B}.$$

The non-strict inequality can be easily proved by means of spherical trigonometry. Consider the following vectors of unit length in $\mathbb R^3$. Let $u=(a,b,c)/\sqrt{A}$, $v=(x,y,z)/\sqrt{B}$, and $w=(1,1,1)/\sqrt{3}$. Let $\alpha=\angle (v,u)$, $\beta=\angle(u,w)$, and $\gamma=\angle(w,v)$. Since $a,b,$ and $c$ are positive, we can easily check that $\cos\beta=(a+b+c)/\sqrt{3A}=(a+b+c)/\sqrt{3(a^2+b^2+c^2)}\ge 1/\sqrt{3}$. Then $\tan\beta\le\sqrt{2}$. By spherical law of cosines, we have that $\cos\gamma=\cos\alpha\cos\beta+\sin\alpha\sin\beta\cos\angle u$. Then $\cos\gamma/\cos\beta\le \cos\alpha+\sin\alpha\tan\beta\le\cos\alpha+\sqrt{2}\sin\alpha$. Substituting the cosines by the inner products, we obtain

$$\frac {(x+y+z)/\sqrt{3B}}{(a+b+c)/\sqrt{3A}}\le\frac{C}{\sqrt{AB}}+\sqrt{2-2\left(\frac{C}{\sqrt{AB}}\right)^2}.$$

The equality should hold only if $\cos\angle u=1$ (that is, the vectors $u,v,w$ and $0$ are coplanar) and $\cos\beta=1/\sqrt{3}$, that is one of $a,b,$ and $c$ is equal to $1$, and the others are equal to $0$.

From the another law of cosines $\cos\beta=\cos\alpha\cos\gamma+\sin\alpha\sin\gamma\cos\angle v$, we can similarly obtain the inequality $$\frac{a+b+c}{x+y+z}\le \frac{C+\sqrt{2(AB-C^2)}}{B}.$$

PS. Maybe the $n$-dimensional version of the inequality holds, if we change in it "$2$" to "$n-1$".

Alex Ravsky
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