Chứng minh : a2+b2+c2<2ab+2bc+2ca
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a: \(VT=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)
\(=a^2c^2+a^2d^2+b^2d^2+b^2c^2\)
\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)
\(=\left(c^2+d^2\right)\left(a^2+b^2\right)\)
b: Bạn ghi lại đề đi bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+b^2d^2-2abcd+a^2d^2-2abcd+b^2c^2\)
\(=a^2c^2+a^2d^2+b^2d^2+b^2c^2\)
\(=\left(c^2+d^2\right)\left(a^2+b^2\right)\)
b: \(\left(ac+bd\right)^2< =\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2c^2+2abcd+b^2d^2-a^2c^2-a^2d^2-b^2c^2-b^2d^2< =0\)
\(\Leftrightarrow-a^2d^2+2abcd-b^2c^2< =0\)
\(\Leftrightarrow\left(ad-bc\right)^2>=0\)(luôn đúng)
a) \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+2abcd+b^2d^2+a^2d^2-2adbc+b^2c^2\)
\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=\left(a^2c^2+a^2d^2\right)+\left(b^2d^2+b^2c^2\right)\)
\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
b) \(\left(a^2+b^2\right)\left(c^2+d^2\right)-\left(ac+bd\right)^{^2}\)
\(=a^2c^2+a^2d^2+b^2c^2+b^2d^2-a^2c^2-2abcd-b^2d^2\)
\(=a^2d^2+b^2c^2-2abcd\)
\(=\left(ad\right)^2-2ad.bc+\left(bc\right)^2\)
\(=\left(ad-bc\right)^2\ge0\)
\(=\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\left(ac+bd\right)^2+\left(ad-bc\right)^2\\ =a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\\ =a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ =\left(a^2c^2+a^2d^2\right)+\left(b^2d^2+b^2c^2\right)\\ =a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\\ =\left(a^2+b^2\right)\left(c^2+d^2\right)\)
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 \(\Leftrightarrow bc=ad\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
\(1\)/
⇔ \(\left(ac\right)^2+2abcd+\left(bd\right)^2+\left(ad\right)^2-2abcd+\left(bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
⇔\(a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
⇔\(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(a^2+b^2\right)\left(c^2+d^2\right)\) ⇒ \(\left(dpcm\right)\)
\(2\)/
⇔\(\left(ac\right)^2+\left(ad\right)^2+\left(bc\right)^2+\left(bd\right)^2\ge\left(ac\right)^2+2abcd+\left(bd\right)^2\)
⇔\(\left(ad\right)^2-2abcd+\left(bc\right)^2\ge0\)
⇔\(\left(ad-bc\right)^2\ge0\left(đúng\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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.
Thêm điều kiện: a,b,c thỏa mãn là các cạnh của một tam giác
Ta có: \(a< b+c\)
nên \(a^2< ab+ac\)
Ta có: b<a+c
nên \(b^2< ab+bc\)
Ta có: c<a+b
nên \(c^2< ac+bc\)
Do đó: \(a^2+b^2+c^2< 2\left(ab+bc+ac\right)\)