Prove the inequality $ax+by+cz+\sqrt{\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)}\ge\frac23(a+b+c)(x+y+z)$

Question

Prove the inequality $$ax+by+cz+\sqrt{\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)}\ge\frac23(a+b+c)(x+y+z)$$ $a,b,c,x,y,z \in \mathbb R$

My attempts:

$$ax+by+cz+\sqrt{\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)}\ge\frac23(a+b+c)(x+y+z)$$

$$(a+x-b-y)^2+(c+z-a-x)^2+(b+y-c-z)^2+(a+b+c-x-y-z)^2 \ge \\ \ge 3 \left( \sqrt{a^2+b^2+c^2}-\sqrt{x^2+y^2+z^2} \right)^2$$ Let $a-x=A, \:b-y=B, \:c-z=C,\:a+x=X, \:b+y=Y, \:c+z=Z$. Then $$(X-Y)^2+(X-Z)^2+(Y-Z)^2+(A+B+C)^2 \ge \\ \ge \frac{3}{4} \left( \sqrt{(A+X)^2+(B+Y)^2+(C+Z)^2}-\sqrt{(A-X)^2+(B-Y)^2+(C-Z)^2} \right)^2$$

Answer 1

Suppose you had two vectors on the unit sphere $\hat u=\langle0,0,1\rangle$ and $\hat v\langle\sin\psi,0,\cos\psi\rangle$. $0\le\psi\le\pi$ and you wanted to find the vector $\hat w=\langle\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta\rangle$ such that $(\hat w\cdot\hat u)(\hat w\cdot\hat v)$ is maximized. Then we want to maximize $$(\sin\theta\cos\phi\sin\psi+\cos\theta\cos\psi)\cos\theta=\frac12\cos\phi\sin\psi\sin2\theta+\frac12\cos\psi(1+\cos2\theta)$$ If $\sin2\theta<0$ we want $\cos\phi=-1$ whereas if $\sin2\theta>0$, $\cos\phi=1$. In either case $(\hat w\cdot\hat u)(\hat w\cdot\hat v)=\frac12\cos\psi+\frac12\cos(2\theta\pm\psi)$ which has a maximum of $$\frac12(1+\cos\psi)=\cos^2\frac{\psi}2$$ By rotations we could have started out with any $2$ unit vectors $\hat u$ and $\hat v$ and found that $$(\hat w\cdot\hat u)(\hat w\cdot\hat v)\le\frac12(1+\hat u\cdot\hat v)$$ Here our $$\begin{align}\hat u&=\frac{\langle a,b,c\rangle}{\sqrt{a^2+b^2+c^2}}\\ \hat v&=\frac{\langle x,y,z\rangle}{\sqrt{x^2+y^2+z^2}}\\ \hat w&=\frac{\langle1,1,1\rangle}{\sqrt{3}}\end{align}$$ So we have $$\frac{(a+b+c)(x+y+z)}{3\sqrt{a^2+b^2+c^2}\sqrt{x^2+y^2+z^2}}\le\frac12\left(1+\frac{ax+by+cz}{\sqrt{a^2+b^2+c^2}\sqrt{x^2+y^2+z^2}}\right)$$ Which works out to $$\frac23(a+b+c)(x+y+z)\le\sqrt{a^2+b^2+c^2}\sqrt{x^2+y^2+z^2}+ax+by+cz$$ I feel kinda bad about the first part of my proof, though: shouldn't there be a more elegant way to show that the unit vector $\hat w$ has to lie halfway between $\hat u$ and $\hat v$?

EDIT: I felt so bad about the first part of my proof that I wanted to show how it could be done cleaner. Again, given $2$ unit vectors $\hat u$ and $\hat v$ we are trying to find the unit vector $\hat w$ that maximizes $\left(\hat w\cdot\hat u\right)\cdot\left(\hat w\cdot\hat v\right)$. We decompose $\hat w$ into its components parallel and perpendicular to the plane containing $\hat u$ and $\hat v$: $$\hat w=\vec w_{\parallel}+\vec w_{\bot}$$ Then $$1=\|\hat w\|^2=\|\vec w_{\parallel}\|^2+2\vec w_{\parallel}\cdot\vec w_{\bot}+\|\vec w_{\bot}\|^2=\|\vec w_{\parallel}\|^2+\|\vec w_{\bot}\|^2$$ So $\|\vec w_{\parallel}\|^2\le1$. Then assuming we can attain a positive value of our objective function $$\left(\hat w\cdot\hat u\right)\cdot\left(\hat w\cdot\hat v\right)=\left(\vec w_{\parallel}\cdot\hat u\right)\cdot\left(\vec w_{\parallel}\cdot\hat v\right)\le\frac{\left(\vec w_{\parallel}\cdot\hat u\right)\cdot\left(\vec w_{\parallel}\cdot\hat v\right)}{\|\vec w_{\parallel}\|^2}$$ So the maximum is attained only for a unit vector $\hat w$ lying in the plane containing $\hat u$ and $\hat v$. This reduces to a $2$-dimensional problem. Let $\hat w\cdot\hat u=\cos\alpha$, $\hat w\cdot\hat v=\cos\xi$, and $\hat u\cdot\hat v=\cos\psi$. Then $$\begin{align}\left(\hat w\cdot\hat u\right)\cdot\left(\hat w\cdot\hat v\right)&=\cos\alpha\cos\xi=\frac12\left(\cos(\alpha+\xi)+\cos(\alpha-\xi)\right)\\ &=\frac12\left(\cos\psi+\cos(\alpha-\xi)\right)\le\frac12\left(\cos\psi+1\right)\end{align}$$ This is familiar from optics where if you have light passing through $2$ plane polarizing filters at angle $\psi$ between them you get maximum transmission when a third such filter at angle $\frac{\psi}2$ is placed between them so the transmission coefficient through the last $2$ filters becomes $\cos^4\frac{\psi}2=\frac14(1+\cos\psi)^2$.

Well, that might have seemed more long-winded than my previous proof and it was, but the cool thing about it is that no reference was made to spherical coordinates so now we can let $$\begin{align}\hat u=\frac{\sum\limits_{i=1}^na_i\hat e_i}{\sqrt{\sum\limits_{i=1}^na_i^2}},&& \hat v=\frac{\sum\limits_{i=1}^nx_i\hat e_i}{\sqrt{\sum\limits_{i=1}^nx_i^2}},&& \hat w=\frac{\sum\limits_{i=1}^n\hat e_i}{\sqrt n}\end{align}$$ And this will work out to $$\sum_{i=1}^na_ix_i+\sqrt{\sum\limits_{i=1}^na_i^2}\sqrt{\sum\limits_{i=1}^nx_i^2}\ge\frac2n\left(\sum_{i=1}^na_i\right)\left(\sum_{i=1}^nx_i\right)$$ Valid for $n\ge1$.

Answer 2

Hint: Since we know that $(a-b)^2+(b-c)^2+(c-a)^2 \ge 0 \implies$ $a^2+b^2+c^2 \ge ab + bc + ac$. Multiplying both sides by 2, we get: $$2(a^2+b^2+c^2) \ge 2(ab + bc + ac)$$ Summing both sides by $a^2 + b^2 + c^2$, we get: $$2(a^2+b^2+c^2) + a^2 + b^2 + c^2 \ge 2(ab + bc + ac) + a^2 + b^2 + c^2$$ $$\implies 3(a^2+b^2+c^2) \ge (a+b+c)^2$$ $$\implies \sqrt{3(a^2+b^2+c^2)} \ge a + b +c$$ Therefore for $x,y$ and $z$, we get: $$\sqrt{3(x^2+y^2+z^2)} \ge x + y +z$$ Multiplying both sides $(1)$: $$3\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)} \ge (a+b+c)(x+y+z)$$ $$\implies 2\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)} \ge \frac{2}{3}(a+b+c)(x+y+z)$$ Since we know that $\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)} \ge ax+by+cz$. Therefore $(2)$: $$3\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)} \ge \frac{2}{3}(a+b+c)(x+y+z)+ax+by+cz$$ Substracting $(1)-(2)$: $$\implies ax + by +cz \ge \frac{1}{3}(a+b+c)(x+y+z)$$ Therefore: $$\frac{1}{3}(a+b+c)(x+y+z)+\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)} \ge \frac{2}{3}(a+b+c)(x+y+z)$$ $$\implies ax + by +cz + \sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)} \ge \frac{2}{3}(a+b+c)(x+y+z)$$

Answer 3

By C-S $$ax+by+cz+\sqrt{\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)}=$$ $$=ax+by+cz+\frac{1}{3}\sqrt{\left(a^2+b^2+c^2\right)\left(9(x^2+y^2+z^2)\right)}=$$ $$=ax+by+cz+\frac{1}{3}\sqrt{\left(a^2+b^2+c^2\right)\sum_{cyc}(2y+2z-x)^2}\geq$$ $$=ax+by+cz+\frac{1}{3}(a(2y+2z-x)+b(2x+2z-y)+c(2x+2y-z))=$$ $$=ax+by+cz+\tfrac{1}{3}(a(2y+2z+2x-3x)+b(2x+2z+2y-3y)+c(2x+2y+2z-3z))=$$ $$=\frac{2}{3}(a+b+c)(x+y+z).$$