6
$\begingroup$

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

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 $$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})\ge (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2$$

  • 0
    Isn't it enough to apply Lagrange's identity for the sum of three squares twice? – 2017-02-20

1 Answers 1

1

We have the following identity. $$(a^2+y^2+z^2)(b^2+x^2+z^2)(c^2+x^2+y^2)-(x^2+y^2+z^2)(ax+by+cz)^2-$$ $$-\frac{1}{2}(ayz-bxz)^2-\frac{1}{2}(ayz-cxy)^2-\frac{1}{2}(bxz-cxy)^2=$$ $$=(x^2+y^2)(x^2+z^2)(y^2+z^2)+a^2b^2c^2+a^2b^2(x^2+y^2)+a^2c^2(x^2+z^2)+b^2c^2(y^2+z^2)-$$ $$-2abxy(x^2+y^2)-2acxz(x^2+z^2)-2bcyz(y^2+z^2)-xyz(abz+acy+bcx).$$ I hope it can help.

  • 0
    Then How to prove the last step $\ge 0?$ – 2017-02-21
  • 0
    @geromty I got something, but it not gives a proof. – 2017-10-05