Lời giải :
\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2\ge a^2x^2+2abxy+b^2y^2\)
\(\Leftrightarrow a^2y^2-2abxy+b^2x^2\ge0\)
\(\Leftrightarrow\left(ay-bx\right)^2\ge0\)( luôn đúng )
Dấu "=" xảy ra \(\Leftrightarrow\frac{a}{x}=\frac{b}{y}\)
\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\) \(\ge\left(ax+by+cz\right)^2\)
Cm bt trên đúng
CM các BĐT
\(a,\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
\(b,\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2\)
\(\)
Chứng minh rằng : \(\left(ax+by\right)^2\ge\left(a^2+b^2\right)\left(x^2+y^2\right)\)
Với mọi a, b, x, y
CM CÁC BẤT ĐẲNG THỨC SAU
A) \(\left(AX+BY\right)^2\le\left(A^2+B^2\right)\left(X^2+Y^2\right)\)
B) \(\left(AX+BY+CZ\right)^2\le\left(A^2+B^2+C^2\right)\left(X^2+Y^2+Z^2\right)\)
Cm\(\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)
CM \(\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)
\(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\ge2x\)Y
Cm Bt trên đúng
CM CÁC HẰNG ĐẲNG THỨC ;
\(\left(A^2+B^2+C^2\right)\left(X^2+Y^2+Z^2\right)=\left(AX+BY+CZ\right)^2+\left(AY-BX\right)^2+\left(AZ-CX\right)^2+\left(BZ-CY\right)^2\)
a, \(\left(a+b\right)^2=2\left(a^2+b^2\right).CM:a=b\)
b,\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2.CM:\frac{a}{x}=\frac{b}{y}\)
c,\(x+y=a+b;x^2+y^2=a^2+b^2.CM:a^3+b^3=x^3+y^3\)
