\(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\)

vì \(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6\le0,\forall x\inℝ\)

\(\Rightarrow B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\le3\)

Dấu "=" xảy ra khi và chỉ khi

\(\dfrac{4}{9}x-\dfrac{2}{15}=0\Rightarrow\dfrac{4}{9}x=\dfrac{2}{15}\Rightarrow x=\dfrac{9}{15}\)

Vậy \(GTLN\left(B\right)=3\left(tạix=\dfrac{9}{15}\right)\)

\(A=\left(2x+\dfrac{1}{3}\right)^4-1\)

vì \(\left(2x+\dfrac{1}{3}\right)^4\ge0,\forall x\inℝ\)

\(\Rightarrow A=\left(2x+\dfrac{1}{3}\right)^4-1\ge-1\)

Dấu "=" xảy ra khi và chỉ khi

\(2x+\dfrac{1}{3}=0\Rightarrow2x=-\dfrac{1}{3}\Rightarrow x=-\dfrac{1}{6}\)

\(\Rightarrow GTNN\left(A\right)=-1\left(tạix=-\dfrac{1}{6}\right)\)

\(P-\dfrac{2}{3}=\dfrac{x^2-6x+9}{3x^2}=\dfrac{\left(x-3\right)^2}{3x^2}\ge0\Rightarrow P\ge\dfrac{2}{3}\).

Dấu "=" xảy ra khi x  =  3.

\(A=2030+\dfrac{8}{x}+\dfrac{1}{x^2}=\left(\dfrac{1}{x}\right)^2+8.\dfrac{1}{x}+16+2014\)

\(\Rightarrow A=\left(\dfrac{1}{x}+4\right)^2+2014\ge2014\)

\(\Rightarrow A_{min}=2014\) khi \(\dfrac{1}{x}+4=0\Rightarrow x=-\dfrac{1}{4}\)

\(A=\dfrac{2030x^2+8x+1}{x^2}\\ =\dfrac{2030x^2}{x^2}+\dfrac{8x}{x^2}+\dfrac{1}{x^2}\\ =2030+\dfrac{8}{x}+\dfrac{1}{x^2}\\ =\left(\dfrac{1}{x}\right)^2+2\cdot\dfrac{1}{x}\cdot4+16+2014\\ =\left(\dfrac{1}{x}+4\right)^2+2014\)

Do \(\left(\dfrac{1}{x}+4\right)^2\ge0,2014>0\)

\(\Rightarrow\left(\dfrac{1}{x}+4\right)^2+2014\ge2014\)

\(\Rightarrow Min\left(A\right)=2014\Leftrightarrow\dfrac{1}{x}+4=0\Rightarrow x=\dfrac{-1}{4}\)

Tử, mẫu không đồng bậc

Đề sai hoặc thiếu điều kiện

tử cộng thêm c^2 bớts c^2

tách tử theo mẫu

cô si mẫu

Đặt \(\dfrac{b}{a}=x;\dfrac{c}{b}=y\).

Ta có: \(P=\dfrac{1}{\left(\dfrac{a+b}{a}\right)^2}+\dfrac{1}{\left(\dfrac{b+c}{b}\right)^2}+\dfrac{b}{a}.\dfrac{c}{b}.\dfrac{1}{4}\)

\(P=\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{\left(y+1\right)^2}+\dfrac{xy}{4}\).

Ta có bđt quen thuộc: \(\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{\left(y+1\right)^2}\ge\dfrac{1}{xy+1}\) (bạn xem cm ở đây).

Do đó \(P\ge\dfrac{1}{xy+1}+\dfrac{xy+1}{4}-\dfrac{1}{4}\ge1-\dfrac{1}{4}=\dfrac{3}{4}\).

Đẳng thức xảy ra khi x = y = 1 tức a = b = c. 

Vậy...

áp dụng bất đẳng thức: 1+b²>=2b. tương tự.....

ad bđt cauchy: a/b+b/c+c/a>=3∛a/b.b/c.c/a=3

P>=\(\dfrac{2ab}{bc}\)+\(\dfrac{2bc}{ca}\)+\(\dfrac{2ca}{ab}\) =2(\(\dfrac{a}{b}\)+\(\dfrac{b}{c}\)+ \(\dfrac{c}{a}\))>=2.3=6

Pmin khi a=b=c=1

Áp dụng bđt : \(1+b^2>=2b\)

bđt cauchy : \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}>3\sqrt[3]{}\) a\b . b\c . c\a = 3

Lời giải:

Đặt $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=t$

$\Rightarrow x=at; y=bt; z=ct$. Ta có:

$(x+y+z)^2=(at+bt+ct)^2=t^2(a+b+c)^2=t^2(*)$

Mặt khác:

$x^2+y^2+z^2=(at)^2+(bt)^2+(ct)^2=t^2(a^2+b^2+c^2)=t^2(**)$

Từ $(*); (**)\Rightarrow (x+y+z)^2=x^2+y^2+z^2$ (đpcm)