\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}\left(đpcm\right)\)

Thay b² = ac vào biểu thức trên, ta có:

\(\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}\)

\(\Rightarrow\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)

\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\)

Áp dụng t/c dtsbn:

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

Ta có \(b^2=ac\Leftrightarrow\dfrac{ac}{c^2}=\dfrac{b^2}{c^2}\Leftrightarrow\dfrac{a}{c}=\dfrac{b^2}{c^2}\left(2\right)\)

\(\left(1\right)\left(2\right)\LeftrightarrowĐpcm\)

Trong tam giác ABC, theo Hệ quả định lý Cô sin ta luôn có :

Mà ta có 2.bc > 0 nên cos A luôn cùng dấu với b2 + c2 – a2.

a) Góc A nhọn ⇔ cos A > 0 ⇔ b2 + c2 – a2 > 0 ⇔ a2 < b2 + c2.

b) Góc A tù ⇔ cos A < 0 ⇔ b2 + c2 – a2 < 0 ⇔ a2 > b2 + c2.

c) Góc A vuông ⇔ cos A = 0 ⇔ b2 + c2 – a2 = 0 ⇔ a2 = b2 + c2.

Nếu \(c^2+d^2\ge1\left(bất.đẳng.thức.đúng\right)\)

Ta chứng minh c²+d²<1

+Đặt x=1-a²-b² và y =1-c² - d²

-0 \(\le x,y\le1\)

Bđt <=> (2 - 2ac - 2bd)²\(\ge\) 4xy <=> ((a-c)²+(b-d)²+x+y)²\(\ge4xy\)

=> ((a-c)²+(b-d)² + x + y)² \(\ge\left(x+y\right)^2\ge4xy\left(đpcm\right)\)

\(\dfrac{a}{b}=\dfrac{b}{c}\Rightarrow ac=b^2\)

\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}\)

Đề thiếu rồi bạn

Ta có

$$a^2+b^2+c^2-ab-bc-ca=0,$$

hay $$\dfrac{1}{2}\left[(a-b)^2+(b-c)^2 +(c-a)^2\right[ = 0.$$

Mà vế trái luôn không âm \(\forall a,b,c \in \mathbb{R}\), đẳng thức xảy ra khi $a=b=c.$

Vậy ta có điều cần chứng minh.

Ta có: \(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow a=b=c\)

\(1,\left(ac+bd\right)^2+\left(ad-bc\right)^2\\ =a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\\ =a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ =\left(a^2c^2+a^2d^2\right)+\left(b^2d^2+b^2c^2\right)\\ =a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\\ =\left(a^2+b^2\right)\left(c^2+d^2\right)\)

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

\(\Leftrightarrow a^2c^2+b^2c^2+a^2d^2+b^2d^2\ge a^2c^2+2abcd+b^2d^2\)

\(\Leftrightarrow b^2c^2-2abcd+a^2d^2\ge0\)

\(\Leftrightarrow\left(bc-ad\right)^2\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow bc=ad\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\)

\(1\)/ 

⇔ \(\left(ac\right)^2+2abcd+\left(bd\right)^2+\left(ad\right)^2-2abcd+\left(bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)

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

⇔\(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(a^2+b^2\right)\left(c^2+d^2\right)\) ⇒ \(\left(dpcm\right)\)

\(2\)/

⇔\(\left(ac\right)^2+\left(ad\right)^2+\left(bc\right)^2+\left(bd\right)^2\ge\left(ac\right)^2+2abcd+\left(bd\right)^2\)

⇔\(\left(ad\right)^2-2abcd+\left(bc\right)^2\ge0\)

⇔\(\left(ad-bc\right)^2\ge0\left(đúng\right)\)

1/ \((ac + bd)^2 + (ad - bc)^2 = (ac)^2 + (bd)^2 + 2(ac)^2 (bd)^2 + (ad)^2 + (bc)^2 - 2(ad)^2 (bc)^2 \)

                                          \(= (ac)^2 + (bd)^2 + 2(acbd)^2 + (ad)^2 + (bc)^2 - 2(adbc)^2 \)

                                          \(= (ac)^2 + (bd)^2 + (ad)^2 + (bc)^2\)

                                          \(= a^2 c^2 + b^2 c^2 + a^2 d^2 + b^2 d^2\)

                                          \(= (a^2 + b^2)c^2 + (a^2 + b^2)d^2\)

                                          \(= (a^2 + b^2)(c^2 + d^2)\)

➤ \((ac + bd)^2 + (ad - bc)^2 = (a^2 + b^2)(c^2 + d^2)\)

2/ \((a^2 + b^2)(c^2 + d^2) ≥ (ac + bd)^2 \) 

↔ \((ac)^2 + (bc)^2 + (ad)^2 + (bd)^2 ≥ (ac)^2 + (bd)^2 + 2(ac)(bd)\)

↔\( (bc)^2 + (ad)^2 ≥ 2(acbd)\)

↔\( (bc)^2 + (ad)^2 - 2(bcad) ≥ 0\)

↔ \( (bc - ad)^2 ≥ 0 \) với mọi a,b,c và d

➤ \((a^2 + b^2)(c^2 + d^2) ≥ (ac + bd)^2 \) với mọi a,b,c,d

Ta có: a+b+c=0

nên a+b=-c

Ta có: \(a^2-b^2-c^2\)

\(=a^2-\left(b^2+c^2\right)\)

\(=a^2-\left[\left(b+c\right)^2-2bc\right]\)

\(=a^2-\left(b+c\right)^2+2bc\)

\(=\left(a-b-c\right)\left(a+b+c\right)+2bc\)

\(=2bc\)

Ta có: \(b^2-c^2-a^2\)

\(=b^2-\left(c^2+a^2\right)\)

\(=b^2-\left[\left(c+a\right)^2-2ca\right]\)

\(=b^2-\left(c+a\right)^2+2ca\)

\(=\left(b-c-a\right)\left(b+c+a\right)+2ca\)

\(=2ac\)

Ta có: \(c^2-a^2-b^2\)

\(=c^2-\left(a^2+b^2\right)\)

\(=c^2-\left[\left(a+b\right)^2-2ab\right]\)

\(=c^2-\left(a+b\right)^2+2ab\)

\(=\left(c-a-b\right)\left(c+a+b\right)+2ab\)

\(=2ab\)

Ta có: \(M=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)

\(=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)

\(=\dfrac{a^3+b^3+c^3}{2abc}\)

Ta có: \(a^3+b^3+c^3\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-cb+c^2\right)-3ab\left(a+b\right)\)

\(=-3ab\left(a+b\right)\)

Thay \(a^3+b^3+c^3=-3ab\left(a+b\right)\) vào biểu thức \(=\dfrac{a^3+b^3+c^3}{2abc}\), ta được: 

\(M=\dfrac{-3ab\left(a+b\right)}{2abc}=\dfrac{-3\left(a+b\right)}{2c}\)

\(=\dfrac{-3\cdot\left(-c\right)}{2c}=\dfrac{3c}{2c}=\dfrac{3}{2}\)

Vậy: \(M=\dfrac{3}{2}\)