\(p+q=1\Rightarrow q=1-p\)
BĐT cần c/m trở thành:
\(pa^2+\left(1-p\right)b^2-p\left(1-p\right)c^2>0\)
\(\Leftrightarrow p^2c^2+\left(a^2-b^2-c^2\right)p+b^2>0\) (1)
\(\Delta=\left(a^2-b^2-c^2\right)^2-4b^2c^2=\left(a^2-b^2-c^2+2bc\right)\left(a^2-b^2-c^2-2bc\right)\)
\(=\left(a^2-\left(b-c\right)^2\right)\left(a^2-\left(b+c\right)^2\right)\)
\(=\left(a+c-b\right)\left(a+b-c\right)\left(a-b-c\right)\left(a+b+c\right)< 0\) theo BĐT tam giác
\(\Rightarrow\) (1) luôn đúng
Ko xài delta thì biến đổi tương đương (1) xuống bằng cách thêm bớt là được:
\(\left(1\right)\Leftrightarrow p^2c^2+2.\dfrac{a^2-b^2-c^2}{2c}.pc+\left(\dfrac{a^2-b^2-c^2}{2c}\right)^2+b^2-\left(\dfrac{a^2-b^2-c^2}{2c}\right)^2>0\)
\(\Leftrightarrow\left(pc+\dfrac{a^2-b^2-c^2}{2c}\right)^2+\dfrac{4b^2c^2-\left(a^2-b^2-c^2\right)^2}{4c^2}>0\)
\(\Leftrightarrow\left(pc+\dfrac{a^2-b^2-c^2}{2c}\right)^2+\dfrac{\left(2bc+a^2-b^2-c^2\right)\left(2bc-a^2+b^2+c^2\right)}{4c^2}>0\)
\(\Leftrightarrow\Leftrightarrow\left(pc+\dfrac{a^2-b^2-c^2}{2c}\right)^2+\dfrac{\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]}{4c^2}>0\)
\(\Leftrightarrow\Leftrightarrow\left(pc+\dfrac{a^2-b^2-c^2}{2c}\right)^2+\dfrac{\left(a+b-c\right)\left(a+c-b\right)\left(a+b+c\right)\left(b+c-a\right)}{4c^2}>0\) (luôn đúng theo BĐT tam giác)
