a)Bunhia:

\(\left(1+2\right)\left(b^2+2a^2\right)\ge\left(1.b+\sqrt{2}.\sqrt{2}a\right)^2=\left(b+2a\right)^2\)

b)\(ab+bc+ca=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng bđt câu a

=>VT\(\ge\)\(\dfrac{b+2a}{\sqrt{3}ab}+\dfrac{c+2b}{\sqrt{3}bc}+\dfrac{a+2c}{\sqrt{3}ca}\)

\(\Leftrightarrow VT\ge\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{1}{c}+\dfrac{2}{a}=3=VP\)

Tự tìm dấu "="

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a + b + 2a² + 2b² ≥ \(2ab+2a\sqrt{b}+2b\sqrt{a}\)

⇔ a + b + 2a² + 2b² - \(2ab-2a\sqrt{b}-2b\sqrt{a}\) ≥ 0

⇔ a² - 2ab + b² + a² - 2a\(\sqrt{b}+b+b^2-2b\sqrt{a}+a\) ≥ 0

⇔ ( a - b)² + ( a - \(\sqrt{b}\) )² + ( b - \(\sqrt{a}\))² ≥ 0 ( Luôn đúng )

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c}\ge4\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\ge2\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\)

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z\ge1\)

\(P=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\)

\(\Rightarrow P\ge\sqrt{\frac{\left(x+2y\right)^2}{3}}+\sqrt{\frac{\left(y+2z\right)^2}{3}}+\sqrt{\frac{\left(z+2x\right)^2}{3}}\)

\(\Rightarrow P\ge\frac{1}{\sqrt{3}}\left(3x+3y+3z\right)\ge\frac{3}{\sqrt{3}}=\sqrt{3}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\) hay \(a=b=c=3\)

Bạn tham khảo:

Câu hỏi của Phạm Vũ Trí Dũng - Toán lớp 8 | Học trực tuyến

Có: a + b = ab \(\le\frac{\left(a+b\right)^2}{4}\)

=> a + b \(\ge4\)

\(\frac{1}{a^2+2a}+\frac{1}{b^2+2b}+\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)

\(\ge\frac{4}{a^2+b^2+2\left(a+b\right)}+\sqrt{\left(1+ab\right)^2}\)

\(=\frac{4}{a^2+b^2+2ab}+\left(1+a+b\right)=\frac{4}{\left(a+b\right)^2}+\left(a+b\right)+1\)

\(=\frac{4}{\left(a+b\right)^2}+\frac{a+b}{4^2}+\frac{a+b}{4^2}+\frac{7}{8}\left(a+b\right)+1\)

\(\ge3\sqrt[3]{\frac{4}{\left(a+b\right)^2}.\frac{a+b}{4^2}.\frac{a+b}{4^2}}+\frac{7}{8}.4+1=\frac{3}{4}+\frac{7}{2}+1\)

Dấu "=" xảy ra <=> a = b = 2

Đặt \(\left(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\right)=\left(x,y,z\right)\) với x, y, z > 0 thì ta có \(x+y+z=1\).

Đặt biểu thức ở VT là A. Ta có: 

\(A=\sqrt{\dfrac{b^2+2a^2}{a^2b^2}}+\sqrt{\dfrac{c^2+2b^2}{b^2c^2}}+\sqrt{\dfrac{a^2+2c^2}{c^2a^2}}=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\).

Ta có bất đẳng thức \(\sqrt{a_1^2+a_2^2}+\sqrt{a_3^2+a_4^2}\ge\sqrt{\left(a_1+a_3\right)^2+\left(a_2+a_4\right)^2}\).

Đây là bđt Mincopxki cho hai bộ số thực và dễ dàng cm bằng biến đổi tương đương.

Do đó \(A\ge\sqrt{\left(x+y\right)^2+\left(\sqrt{2}y+\sqrt{2}z\right)^2}+\sqrt{z^2+2x^2}\ge\sqrt{\left(x+y+z\right)^2+\left(\sqrt{2}y+\sqrt{2}z+\sqrt{2}x\right)^2}=\sqrt{1+2}=\sqrt{3}=VP\).

Đẳng thức xảy ra khi a = b = c = 3.

Vậy...

Tương tự: \(GT\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

\(VT=\dfrac{\sqrt{a^2+a^2+b^2}}{ab}+\dfrac{\sqrt{b^2+b^2+c^2}}{bc}+\dfrac{\sqrt{c^2+a^2+a^2}}{ca}\)

\(VT\ge\dfrac{\sqrt{\dfrac{1}{3}\left(a+a+b\right)^2}}{ab}+\dfrac{\sqrt{\dfrac{1}{3}\left(b+b+c\right)^2}}{bc}+\dfrac{\sqrt{\dfrac{1}{3}\left(c+c+a\right)^2}}{ca}\)

\(VT\ge\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Vì \(a\in\left[-2;5\right]\Rightarrow\left(a+2\right)\left(a-5\right)\le0\Leftrightarrow a^2-3a-10\le0\Leftrightarrow a^2\le3a+10\)(1)

CMTT \(b^2\le3b+10\Rightarrow2b^2\le6b+20\left(2\right)\) ; \(c^2\le3c+10\Leftrightarrow3c^2\le9c+30\)(3)

        Từ (1) (2) và (3) => \(a^2+2b^2+3c^2\le3\left(a+2b+3c\right)+60\le3.2+60=66\)

BĐT đc cm 

xin lỗi mình mới học lớp  8

Trả lời:

a. Áp dụng BĐT Cô-si: x + y\(\ge\) \(2\sqrt{xy}\) (với x,y\(\ge\)0)

Ta có: a + b\(\ge\)\(2\sqrt{ab}\)

b+c\(\ge\)\(2\sqrt{bc}\)

c+a\(\ge\)\(2\sqrt{ca}\)

\(\Rightarrow\) (a+b)(b+c)(c+a) \(\ge\)\(8\sqrt{a^2b^2c^2}\)= 8abc (đpcm)

b. Áp dụng BĐT Cô-si: \(\sqrt{ab}\)\(\le\)\(\dfrac{a+b}{2}\) ( với a,b\(\ge\)0)

Ta có: \(\sqrt{3a\left(a+2b\right)}\)\(\le\)\(\dfrac{3a+a+2b}{2}\)=\(\dfrac{4a+2b}{2}\)=2a+b

\(\Rightarrow\) \(a\sqrt{3a\left(a+2b\right)}\)\(\le\)a(2a+b) = 2a²+ab

CMTT: \(b\sqrt{3b\left(b+2a\right)}\)\(\le\)b(2b+a) = 2b²+ab

\(\rightarrow\)\(a\sqrt{3a\left(a+2b\right)}\)+\(b\sqrt{3b\left(2b+a\right)}\)\(\le\) 2a²+ab+2b²+ab

= 2(a²+b²)+2ab =6(đpcm)

c. Áp dụng BĐT Cô-si với 3 số a+b; b+c;c+a

Ta có: (a+b)(b+c)(c+a)\(\le\)\(\left(\dfrac{2\left(a+b+c\right)}{3}\right)^3\)

\(\Leftrightarrow\) 1 \(\le\) \(\dfrac{8}{27}\left(a+b+c\right)^3\)

\(\Leftrightarrow\) (a+b+c)³ \(\ge\) \(\dfrac{8}{27}\)

\(\Leftrightarrow\) a+b+c \(\ge\) \(\dfrac{3}{2}\) (1)

Lại có: (a+b)(b+c)(c+a) = (a+b+c)(ab+bc+ca) -abc

\(\Leftrightarrow\) 1= (a+b+c)(ab+bc+ca) - abc

\(\Leftrightarrow\) ab+bc+ca = \(\dfrac{1+abc}{a+b+c}\) (2)

Theo câu a. (a+b)(b+c)(c+a) \(\ge\) 8abc

\(\Leftrightarrow\) 1 \(\ge\) 8abc

\(\Leftrightarrow\) abc \(\le\)\(\dfrac{1}{8}\) (3)

Từ (1),(3) kết hợp với (2)

\(\Rightarrow\) ab+bc+ca \(\le\) \(\dfrac{1+\dfrac{1}{8}}{\dfrac{3}{2}}\) = \(\dfrac{3}{4}\) (đpcm)