9

Following problem I have post MO ,we know Lagrange's identity $$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})=(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2+\sum_{i=1}^{2}\sum_{j=i+1}^{3}(a_{i}b_{j}-a_{j}b_{i})^2$$

then we have Cauchy-Schwarz inequality $$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})\ge (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2$$

However, does the following inequality still hold $$(a^2_{1}+b^2_{2}+b^2_{3})(a^2_{2}+b^2_{3}+b^2_{1})(a^2_{3}+b^2_{1}+b^2_{2})\ge (b^2_{1}+b^2_{2}+b^2_{3})(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2 $$ $$+\dfrac{1}{2}(b_{1}a_{2}b_{3}-b_{1}b_{2}a_{3})^2+\dfrac{1}{2}(b_{1}b_{2}a_{3}-a_{1}b_{2}b_{3})^2+\dfrac{1}{2}(a_{1}b_{2}b_{3}-b_{1}a_{2}b_{3})^2\tag{*}$$

for $a_{i},b_{i}\in \mathbb R,i=1,2,3$?

I think maybe exist simple methods

Community
  • 1
math110
  • 93,304

1 Answers1

0

Your inequality is true .We have to prove this with $a,b,c,x,y,z$ real numbers:

$$(a^2+y^2+z^2)(b^2+x^2+z^2)(c^2+x^2+y^2)-(x^2+y^2+z^2)(ax+by+cz)^2-$$ $$-\frac{1}{2}(ayz-bxz)^2-\frac{1}{2}(ayz-cxy)^2-\frac{1}{2}(bxz-cxy)^2\geq0$$

All the proof is based on the following substitution :

$$u=x+a$$ $$v=x-a\geq 0$$ $$w=y+b$$ $$p=y-b\geq 0$$ $$q=z+c$$ $$r=z-c\geq 0$$

We get :

$$\frac{1}{64} ((u - v)^2 + (w + p)^2 + (q + r)^2) ((u + v)^2 + (w - p)^2 + (q + r)^2) ((u + v)^2 + (w + p)^2 + (q - r)^2) - (0.25 (u^2 - v^2) + 0.25 (w^2 - p^2) + 0.25 (q^2 - r^2))^2 (0.25 (u + v)^2 + 0.25 (p + w)^2 + 0.25 (q + r)^2)\geq 0.5 (\frac{1}{8}(u - v) (q + r) (w + p) - \frac{1}{8} (u + v) (q - r) (w + p))^2 + 0.5 (\frac{1}{8} (u - v) (q + r) (w + p) -\frac{1}{8}(u + v) (q + r) (w - p))^2 + 0.5 (\frac{1}{8} (u + v) (q + r) (w - p) - \frac{1}{8}(u + v) (q - r) (w + p))^2$$

The LHS becomes : $$ 0.0625 (p^4 q^2 + p^4 u^2 + 2. p^3 q^2 w + 2. p^3 q r w + 2. p^3 u^2 w + 2. p^3 u v w + p^2 q^4 + 2. p^2 q^3 r + p^2 q^2 r^2 + \color{yellowgreen}{\text{$2 p^2 q^2 u^2$}} +\color{yellowgreen}{\text{$ 2 p^2 q^2 u v$}} + \color{yellowgreen}{\text{$ 2p^2 q^2 v^2$}} + p^2 q^2 w^2 + \color{yellowgreen}{\text{$ 2 p^2 q r u^2$}} + 2. p^2 q r u v + 4. p^2 q r w^2 + \color{yellowgreen}{\text{$ 2 p^2 r^2 u^2$}} + p^2 r^2 w^2 + p^2 u^4 + 2. p^2 u^3 v + p^2 u^2 v^2 + p^2 u^2 w^2 + 4. p^2 u v w^2 + p^2 v^2 w^2 + 2. p q^3 r w + 4. p q^2 r^2 w + 2. p q^2 u v w + \color{yellowgreen}{\text{$ 2 p q^2 v^2 w$}} + 2. p q r^3 w + 2. p q r u^2 w\color{yellowgreen}{\text{$ - 4 p q r u v w $}} + 2. p q r v^2 w + 2. p q r w^3 + \color{yellowgreen}{\text{$2 p r^2 u^2 w $}} + 2. p r^2 u v w + 2. p r^2 w^3 + 2. p u^3 v w + 4. p u^2 v^2 w + 2. p u v^3 w + 2. p u v w^3 + 2. p v^2 w^3 + q^4 v^2 + 2. q^3 r u v + 2. q^3 r v^2 + q^2 r^2 u^2 + 4. q^2 r^2 u v + q^2 r^2 v^2 + q^2 r^2 w^2 + q^2 u^2 v^2 + 2. q^2 u v^3 + q^2 v^4 + \color{yellowgreen}{\text{$2 q^2 v^2 w^2 $}} + 2. q r^3 u^2 + 2. q r^3 u v + 2. q r^3 w^2 + 2. q r u^3 v + 4. q r u^2 v^2 + 2. q r u v^3 + 2. q r u v w^2 + \color{yellowgreen}{\text{$2q r v^2 w^2 $}} + r^4 u^2 + r^4 w^2 + r^2 u^4 + 2. r^2 u^3 v + r^2 u^2 v^2 + \color{yellowgreen}{\text{$2 r^2 u^2 w^2 + 2r^2 u v w^2$}} + \color{yellowgreen}{\text{$2 r^2 v^2 w^2$}} + r^2 w^4 + u^2 v^2 w^2 + 2. u v^3 w^2 + v^4 w^2 + v^2 w^4)$$

And the RHS :

$$0.0625 (\color{yellowgreen}{\text{$ p^2 q^2 u^2$}} +\color{yellowgreen}{\text{$ p^2 q^2 u v$}} + \color{yellowgreen}{\text{$ p^2 q^2 v^2$}} +\color{yellowgreen}{\text{$ 2 p^2 q r u^2$}} - p^2 q r u v +\color{yellowgreen}{\text{$p^2 r^2 u^2$}} - p q^2 u v w + \color{yellowgreen}{\text{$ p q^2 v^2 w$}} - p q r u^2 w \color{yellowgreen}{\text{$ - 6 p q r u v w $}} - p q r v^2 w + \color{yellowgreen}{\text{$ p r^2 u^2 w $}} - p r^2 u v w + \color{yellowgreen}{\text{$ q^2 v^2 w^2 $}} - q r u v w^2 + \color{yellowgreen}{\text{$q r v^2 w^2 $}} +\color{yellowgreen}{\text{$2 r^2 u^2 w^2 + 2r^2 u v w^2$}} + \color{yellowgreen}{\text{$ r^2 v^2 w^2$}} )$$

I puts colors on the polynomials indicated.And you can compare it in the order .

So we can conclude that the inequality is true in this case .

What's happens if we have :

$$u=x+a$$ $$v=x-a\leq 0$$ $$w=y+b$$ $$p=y-b\leq 0$$ $$q=z+c$$ $$r=z-c\leq 0$$

We have this inequality :

$$\frac{1}{64} ((u - v)^2 + (w + p)^2 + (q + r)^2) ((u + v)^2 + (w - p)^2 + (q + r)^2) ((u + v)^2 + (w + p)^2 + (q - r)^2) - (0.25 (u^2 - v^2) + 0.25 (w^2 - p^2) + 0.25 (q^2 - r^2))^2 (0.25 (u + v)^2 + 0.25 (p + w)^2 + 0.25 (q + r)^2)\geq 0.5 (\frac{1}{8}(u - v) (q + r) (w + p) - \frac{1}{8} (u + v) (q - r) (w + p))^2 + 0.5 (\frac{1}{8} (u - v) (q + r) (w + p) -\frac{1}{8}(u + v) (q + r) (w - p))^2 + 0.5 (\frac{1}{8} (u + v) (q + r) (w - p) - \frac{1}{8}(u + v) (q - r) (w + p))^2$$

So it's changes nothings

The other cases can be treated in the same way.

  • Even if the colored polynomials from both sides cancel and there remains only polynomials with “$+$” at the left hand side and only polynomials with “$-$” at the right hand side it does not prove the inequality, because not all of the variables $u,v,w,p,q,r$ are non-negative. Even if we force to that some of them and, there still remains some polynomials at the right hand side which are not necessarily non-negative, for instance, $-p^2qruv$. Especially taking into account that we treat the other cases in the same way, that is the choice of signs is not essential. – Alex Ravsky Oct 22 '17 at 12:55