Lời giải:

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

$\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0$

$\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)=0$

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

Vì $(a-b)^2; (b-c)^2; (c-a)^2\geq 0$ với mọi $a,b,c$ nên để tổng của chúng bằng $0$ thì:

$a-b=b-c=c-a=0$

$\Rightarrow a=b=c$

$\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1$

Khi đó:

$(\frac{a}{b}+1)(\frac{b}{c}+1)(\frac{c}{a}+1)=(1+1)(1+1)(1+1)=8$ 

Ta có đpcm.

\(P=\dfrac{a^2+b^2+c^2}{ab+bc+ca}\ge\dfrac{ab+bc+ca}{ab+bc+ca}=1\)

\(P_{min}=1\) khi \(a=b=c=1\)

\(P=\dfrac{\left(a+b+c\right)^2-2\left(ab+bc+ca\right)}{ab+bc+ca}=\dfrac{9}{ab+bc+ca}-2\)

Do \(a;b\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab\ge a+b-1=2-c\)

\(\Rightarrow ab+c\left(a+b\right)\ge2-c+c\left(3-c\right)=-c^2+2c+2=c\left(2-c\right)+2\ge2\)

\(\Rightarrow P\le\dfrac{9}{2}-2=\dfrac{5}{2}\)

\(P_{max}=\dfrac{5}{2}\) khi \(\left(a;b;c\right)=\left(1;2;0\right);\left(2;1;0\right)\)

86 vì ta học lớp 9

Ta có: \(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)\)

\(=a\left(b^2c^2-b^2-c^2+1\right)+b\left(a^2c^2-a^2-c^2+1\right)\)

\(+c\left(a^2b^2-a^2-b^2+1\right)\)

\(=ab^2c^2-ab^2-ac^2+a+ba^2c^2-a^2b-bc^2+b\)

\(+ca^2b^2-a^2c-b^2c+c\)

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

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

\(=abc\left(bc+ac+ab\right)+\left(a+b+c\right)\)\(-\left[ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\right]\)

\(=abc\left(bc+ac+ab\right)+\left(a+b+c\right)+3abc\)\(-\left[ab\left(a+b+c\right)+bc\left(a+b+c\right)+ca\left(a+b+c\right)\right]\)

\(=abc\left(bc+ac+ab\right)+\left(a+b+c\right)+3abc\)\(-\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(=abc\left(bc+ac+ab\right)+abc+3abc\)\(-abc\left(ab+bc+ca\right)=4abc\)

Vậy \(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)=4abc\)(đpcm)

Ta chứng minh BĐT sau cho các số dương:

\(x^5+y^5\ge xy\left(x^3+y^3\right)\)

\(\Leftrightarrow x^5-x^4y+y^5-xy^4\ge0\)

\(\Leftrightarrow\left(x^4-y^4\right)\left(x-y\right)\ge0\)

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

Áp dụng:

\(\dfrac{a^5+b^5}{ab\left(a+b\right)}\ge\dfrac{ab\left(a^3+b^3\right)}{ab\left(a+b\right)}=\dfrac{a^3+b^3}{a+b}=a^2-ab+b^2\)

Tương tự và cộng lại:

\(VT\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)=2-\left(ab+ca+ca\right)\)

\(VT\ge4-\left(ab+bc+ca\right)-2=4\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)-2\)

\(VT\ge4\left(ab+bc+ca\right)-\left(ab+bc+ca\right)-2=3\left(ab+bc+ca\right)-2\) (đpcm)

\(a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)

Ta có: \(a^2+b^2+c^2=ab+bc+ca\)

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

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

\(\Leftrightarrow a=b=c\)

⇔⎧⎪⎨⎪⎩a−b=0b−c=0c−a=0⇔{a−b=0b−c=0c−a=0 

a^2 hay a.2 thế

a^2 bn ạ!!
 

Câu hỏi của Hattory Heiji - Toán lớp 8 - Học toán với OnlineMath

tvbobnokb' n

iai

  ni;bv nn0

Sửa đề: 1+a^2;1+b^2;1+c^2

\(\dfrac{a}{\sqrt{1+a^2}}=\dfrac{a}{\sqrt{a^2+ab+c+ac}}=\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}< =\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)

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

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

=>\(A< =\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{3}{2}\)