§1. Bất đẳng thức

Nguyễn Việt Lâm
10 tháng 4 2022 lúc 22:16

1. Tìm min P:

Cách 1: nhanh nhất là chúng ta sử dụng BĐT Holder:

\(\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)\left(a^2b^2+b^2c^2+c^2a^2\right)\ge\left(a^2+b^2+c^2\right)^3\)

\(=\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)^2\ge\left(a^2+b^2+c^2\right).3\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow P^2\ge3\left(a^2+b^2+c^2\right)=9\)

\(\Rightarrow P\ge3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

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Nguyễn Việt Lâm
10 tháng 4 2022 lúc 22:19

Nếu không sử dụng Holder, ta làm như sau:

\(P^2=\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)^2=\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}+\dfrac{c^4}{a^2}+\dfrac{2a^2b}{c}+\dfrac{2b^2c}{a}+\dfrac{2c^2a}{b}\)

\(P^2=\left(\dfrac{a^4}{b^2}+\dfrac{a^2b}{c}+\dfrac{a^2b}{c}+c^2\right)+\left(\dfrac{b^4}{c^2}+\dfrac{b^2c}{a}+\dfrac{b^2c}{a}+a^2\right)+\left(\dfrac{c^4}{a^2}+\dfrac{c^2a}{b}+\dfrac{c^2a}{b}+b^2\right)-3\)

\(P^2\ge4\sqrt[4]{\dfrac{a^8b^2c^2}{b^2c^2}}+4\sqrt[4]{\dfrac{b^8a^2c^2}{a^2c^2}}+4\sqrt[4]{\dfrac{c^8a^2b^2}{a^2b^2}}-3\)

\(P^2\ge4\left(a^2+b^2+c^2\right)-3=9\)

\(\Rightarrow P\ge3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

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Nguyễn Việt Lâm
10 tháng 4 2022 lúc 22:28

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

Ta có:

\(a^3+1+1\ge3a\Leftrightarrow2\ge a\left(3-a^2\right)\)

\(\Rightarrow\dfrac{1}{3-a^2}\ge\dfrac{a}{2}\Rightarrow\dfrac{a}{3-a^2}\ge\dfrac{a^2}{2}\)

Tương tự:

\(\dfrac{b}{3-b^2}\ge\dfrac{b^2}{2}\) ; \(\dfrac{c}{3-c^2}\ge\dfrac{c^2}{2}\)

Cộng vế:

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

Dấu "=" xảy ra khi \(a=b=c=1\)

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Xu 6 xí=))
10 tháng 4 2022 lúc 16:37

lỗi

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Mạnh=_=
10 tháng 4 2022 lúc 16:37

looix

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tuechi ヾ(•ω•`)o🖤🖤🖤🖤
10 tháng 4 2022 lúc 16:37

lỗi cj ạ

 

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Nguyễn Việt Lâm
8 tháng 4 2022 lúc 13:25

\(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=1\)

\(\Rightarrow\dfrac{a}{a+1}=\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge2\sqrt{\dfrac{1}{\left(b+1\right)\left(c+1\right)}}\)

Tương tự: \(\dfrac{b}{b+1}\ge2\sqrt{\dfrac{1}{\left(a+1\right)\left(c+1\right)}}\) ; \(\dfrac{c}{c+1}\ge2\sqrt{\dfrac{1}{\left(a+1\right)\left(b+1\right)}}\)

Nhân vế:

\(\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{8}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)

\(\Rightarrow abc\ge8\)

a.

\(P=a\sqrt{4+b^2}+b\sqrt{4+c^2}+c\sqrt{4+a^2}\)

(Áp dụng \(4+x^2\ge\dfrac{1}{2}\left(2+x\right)^2\))

\(P\ge\dfrac{a}{\sqrt{2}}\left(2+b\right)+\dfrac{b}{\sqrt{2}}\left(2+c\right)+\dfrac{c}{\sqrt{2}}\left(2+a\right)\)

\(P\ge\dfrac{1}{\sqrt{2}}\left(2a+2b+2c+ab+bc+ca\right)\ge\dfrac{1}{\sqrt{2}}.\left(2.3\sqrt[3]{abc}+3\sqrt[3]{\left(abc\right)^2}\right)\ge12\sqrt{2}\)

Dấu "=" xảy ra khi  \(a=b=c=2\)

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Nguyễn Việt Lâm
8 tháng 4 2022 lúc 13:26

b.

\(\dfrac{1}{\sqrt{a^3+1}}=\dfrac{1}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\ge\dfrac{2}{a+1+a^2-a+1}=\dfrac{2}{a^2+2}\)

Tương tự: \(\dfrac{1}{\sqrt{b^3+1}}\ge\dfrac{2}{b^2+2}\) ; \(\dfrac{1}{\sqrt{c^3+1}}\ge\dfrac{2}{c^2+2}\)

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

Từ điều kiện ban đầu \(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=1\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{1+a}=x\\\dfrac{1}{1+b}=y\\\dfrac{1}{1+c}=z\end{matrix}\right.\) \(\Rightarrow x+y+z=1\)

Đồng thời: \(\dfrac{1}{1+a}=x\Rightarrow1+a=\dfrac{1}{x}=\dfrac{x+y+z}{x}\Rightarrow a=\dfrac{x+y+z}{x}-1=\dfrac{y+z}{x}\)

Tương tự: \(b=\dfrac{z+x}{y}\) ; \(c=\dfrac{x+y}{z}\)

Từ đó:

\(Q\ge\dfrac{2}{\left(\dfrac{y+z}{x}\right)^2+2}+\dfrac{2}{\left(\dfrac{z+x}{y}\right)^2+2}+\dfrac{2}{\left(\dfrac{x+y}{z}\right)^2+2}\)

\(Q\ge\dfrac{2x^2}{2x^2+\left(y+z\right)^2}+\dfrac{2y^2}{2y^2+\left(z+x\right)^2}+\dfrac{2z^2}{2z^2+\left(x+y\right)^2}\)

\(Q\ge\dfrac{2x^2}{2x^2+2\left(y^2+z^2\right)}+\dfrac{2y^2}{2y^2+2\left(z^2+x^2\right)}+\dfrac{2z^2}{2z^2+2\left(x^2+y^2\right)}=1\)

\(Q_{min}=1\) khi \(x=y=z\) hay \(a=b=c=2\)

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wibu
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Phạm Kim Oanh
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Nguyễn Việt Lâm
6 tháng 4 2022 lúc 17:30

Ta có:

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

\(\Rightarrow\dfrac{a}{\sqrt{a^2+b+c}}\le\dfrac{a\sqrt{1+b+c}}{a+b+c}\)

Tương tự: \(\dfrac{b}{\sqrt{b^2+a+c}}\le\dfrac{b\sqrt{1+c+a}}{a+b+c}\) ; \(\dfrac{c}{\sqrt{c^2+b+a}}\le\dfrac{c\sqrt{1+a+b}}{a+b+c}\)

Cộng vế:

\(P\le\dfrac{a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}}{a+b+c}\)

Lại có:

\(a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}\)

\(=\sqrt{a}.\sqrt{a+ab+ac}+\sqrt{b}.\sqrt{b+bc+ab}+\sqrt{c}.\sqrt{c+ac+bc}\)

\(\le\sqrt{\left(a+b+c\right)\left(a+b+c+2ab+2bc+2ca\right)}\)

\(\Rightarrow P\le\dfrac{\sqrt{\left(a+b+c\right)\left(a+b+c+2ab+bc+ca\right)}}{a+b+c}=\sqrt{\dfrac{a+b+c+2ab+2bc+2ca}{a+b+c}}\)

Do đó ta chỉ cần chứng minh:

\(\dfrac{a+b+c+2ab+2bc+2ca}{a+b+c}\le3\Leftrightarrow a+b+c\ge ab+bc+ca\)

Thật vậy:

\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)

\(\Rightarrow a+b+c\ge ab+bc+ca\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

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Phạm Kim Oanh
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Nguyễn Việt Lâm
6 tháng 4 2022 lúc 17:15

Ta có:

\(\left(b^2+c^2+1\right)\left(1+1+a^2\right)\ge\left(a+b+c\right)^2=9\)

\(\Rightarrow\dfrac{1}{b^2+c^2+1}\le\dfrac{a^2+2}{9}\)

\(\Rightarrow\dfrac{a}{b^2+c^2+1}\le\dfrac{a^3+2a}{9}\)

Tương tự: \(\dfrac{b}{c^2+a^2+1}\le\dfrac{b^3+2b}{9}\) ; \(\dfrac{c}{a^2+b^2+1}\le\dfrac{c^3+2c}{9}\)

Cộng vế:

\(VT\le\dfrac{a^3+b^3+c^3+2\left(a+b+c\right)}{9}=\dfrac{a^3+b^3+c^3+6}{9}\) (1)

Lại có:

\(\left(a^3+1+1\right)+\left(b^3+1+1\right)+\left(c^3+1+1\right)\ge3a+3b+3c\)

\(\Rightarrow a^3+b^3+c^3\ge3\Rightarrow6\le2\left(a^3+b^3+c^3\right)\) (2)

(1);(2) \(\Rightarrow VT\le\dfrac{a^3+b^3+c^3+2\left(a^3+b^3+c^3\right)}{9}=\dfrac{a^3+b^3+c^3}{3}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

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Phạm Kim Oanh
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Khôi Bùi
2 tháng 4 2022 lúc 7:59

Dễ dàng c/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\)

Ta có : \(\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\dfrac{1}{a+b+4}\le\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}\right)\) 

Suy ra : \(\Sigma\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le2.\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)=\dfrac{1}{2}.1=\dfrac{1}{2}\) 

" = " \(\Leftrightarrow a=b=c=1\)

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Phạm Kim Oanh
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Khôi Bùi
2 tháng 4 2022 lúc 7:35

C/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\) (*)

Thật vậy , (*) \(\Leftrightarrow\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(a+2\right)\left(c+2\right)=\left(a+2\right)\left(b+2\right)\left(c+2\right)\)

\(\Leftrightarrow ab+bc+ac+4\left(a+b+c\right)+12=abc+2\left(ab+bc+ac\right)+4\left(a+b+c\right)+8\)

\(\Leftrightarrow ab+bc+ac+abc=4\) (Đ)

=> (*) đúng ( đpcm ) 

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Phạm Kim Oanh
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Nguyễn Việt Lâm
6 tháng 4 2022 lúc 17:09

\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{ab}{2b}\right)\)

\(=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)

Tương tự:

\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{b}{2}\right)\)

\(\dfrac{ac}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ac}{b+c}+\dfrac{ac}{a+b}+\dfrac{c}{2}\right)\)

Cộng vế:

\(P\le\dfrac{1}{9}\left(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{a+b+c}{2}\right)\)

\(P\le\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

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Nguyễn Vũ Quỳnh Như
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