\(\left|ab+cd\right|\le\sqrt{\left(a^2+c^2\right)\left(b^2+d^2\right)}\Leftrightarrow\left|ab+cd\right|^2\le\left(a^2+c^2\right)\left(b^2+d^2\right)\)

Áp dụng bất đẳng thức bunhiacopxki ta suy ra:

 Dấu "=" xảy ra <=> ad=bc

\(ab+bc+ca\le a^2+b^2+c^2\le\frac{\left(a+b+c\right)^2}{3}\) ( bđt phụ + Cauchy-Schwarz dạng Engel ) 

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)

CM bđt phụ : \(x^2+y^2+z^2\ge xy+yz+zx\)

\(\Leftrightarrow\)\(2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)

\(\Leftrightarrow\)\(2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)

\(\Leftrightarrow\)\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\ge0\)

\(\Leftrightarrow\)\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) ( luôn đúng ) 

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z\)

Chúc bạn học tốt ~ 

Bai 1: Ap dung BDT Bunhiacopxki ta co:

         \(ax+by+cz+2\sqrt {(ab+ac+bc)(xy+yz+xz)} \)

         \(≤ \sqrt {(a^2+b^2+c^2)(x^2+y^2+z^2)} + \sqrt {(ab+ac+bc)(xy+yz+zx)}+\sqrt {(ab+ac+bc)(xy+yz+zx)}\)

         \(≤ \sqrt {(a^2+b^2+c^2+2ab+2ac+2bc)(x^2+y^2+z^2+2xy+2yz+2zx)}\)

         \(= (a+b+c)(x+y+z)\) 

   =>  \(Q.E.D\)

Tiep bai 4:Ta co:

               BDT <=>  \((2+y^2z)(2+z^2x)(2+x^2y)≥(2+x)(2+y)(2+z)\)

    Sau khi khai trien con:   \(2(z^2x+y^2z+x^2y)+x^2z+z^2y+y^2x≥xy+yz+zx+2x+2y+2z \)

               Ap dung BDT Cosi ta co:

                                       \(z^2x+x ≥ 2zx \) <=> \(z^2x≥2zx-x\)

              Lam tuong tu ta co:  \(2(z^2x+y^2z+x^2y)≥4xy+4yz+4zx-2x-2y-2z \)(1)

                                        \(x^2z+{1\over z}≥2x \) <=> \(x^2z≥2x-xy \) (do xyz=1)

              Lam tuong tu ta co:  \(x^2z+z^2y+y^2x≥ 2y+2z+2x-xy-yz-zx\)(2)

Cong (1) voi (2) ta co:      VT\(≥ 3(xy+yz+zx)\)(*)

               Voi cach lam tuong tu ta cung duoc:  VT\(≥ 3(x+y+z) \)(**)

Tu (*) va (**) suy ra :   \(3 \)VT \(≥ 6(x+y+z)+3(xy+yz+zx) \)

                           <=>   VT \(≥ 2(x+y+z)+xy+yz+zx\)

                            =>   \(Q.E.D\)

\(ax+by+cz=0\Rightarrow\left(ax+by+cz\right)^2=0\)

\(\Rightarrow a^2x^2+b^2y^2+c^2z^2=-2\left(axby+bycz+axcz\right)\)

Ta co

\(\dfrac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(z-x\right)^2+ab\left(x-y\right)^2}\)

\(=\dfrac{ax^2+by^2+cz^2}{bcy^2-2bcyz+bcz^2+acz^2-2aczx+acx^2+abx^2-2abxy+aby^2}\)

\(=\dfrac{ax^2+by^2+cz^2}{bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2-2\left(axby+bcyz+axcz\right)}\)

\(=\dfrac{ax^2+by^2+cz^2}{bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2}\)

\(=\dfrac{ax^2+by^2+cz^2}{\left(acx^2+abx^2+a^2x^2\right)+\left(bcy^2+aby^2+b^2y^2\right)+\left(c^2z^2+acz^2+bcz^2\right)}\)

\(=\dfrac{ax^2+by^2+cz^2}{ax^2\left(a+b+c\right)+by^2\left(a+b+c\right)+cz^2\left(a+b+c\right)}\)

\(=\dfrac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\dfrac{1}{a+b+c}\) ( dpcm)

BĐT tương đương

\(\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)\left(a+b+c\right)\ge\left(x+y+z\right)^2\)

sau đó nhân phá ra và đưa về dạng tổng các bình phương

Câu a)

Em mới hc lớp 7 nên chỉ chứng minh cái phần dấu bằng xảy ra khi nào thui. Ko biết có đúng ko

Theo đề bài Ta có

\(\left(ac+bd\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)

\(\Leftrightarrow\left(ac+bd\right)^2=\left(a^2+b^2\right)^2=\left(c^2+d^2\right)^2\)

Suy ra \(ac=a^2,bd=b^2,ac=b^2\)

Suy ra \(a=b=c=d\)

Vậy dấu bằng xảy ra khi \(a=b=c=d\)

ukm nhưng anh cần câu b

\(6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)

\(=6\left(x^2+y^2+z^2\right)+12\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)-2\left(xy+yz+xz\right)\)

\(=6\left(x+y+z\right)^2+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{2z+x+y}\right)-2\left(xy+yz+xz\right)\)

\(\ge6\left(x+y+z\right)^2+2.\dfrac{\left(1+1+1\right)^2}{2x+y+z+x+2y+z+2z+x+y}-2\left(xy+yz+xz\right)\)

\(=6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-2\left(xy+yz+xz\right)\)

\(\ge6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-\dfrac{2}{3}\left(x+y+z\right)^2\)

\(=6.\left(\dfrac{3}{4}\right)^2+\dfrac{18}{4.\dfrac{3}{4}}-\dfrac{2}{3}.\left(\dfrac{3}{4}\right)^2=9\)

\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)

a) ab+bc+ca\(\le\dfrac{\left(a+c+b\right)^2}{3}\)

\(\Leftrightarrow3ab+3bc+3ac\le a^2+b^2+c^2+2ab+2bc+2ac\)

\(\Leftrightarrow ab+bc+ac\le a^2+b^2+c^2\)

\(\Leftrightarrow2ab+2bc+2ca\le2a^2+2b^2+2c^2\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng \(\forall a,b,c\)

\(LHS\ge\left(\sqrt{ax}.\sqrt{\frac{a}{x}}+\sqrt{bx}.\sqrt{\frac{b}{x}}+\sqrt{cx}.\sqrt{\frac{c}{x}}\right)^2=\left(a+b+c\right)^2\)