Chương 4: BẤT ĐẲNG THỨC, BẤT PHƯƠNG TRÌNH

thằng nào tên mịnh khai mau ; còn hiẻu là j z bn

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left (\frac{1}{a}+\frac{1}{b}\right)(a+b)\geq (1+1)^2\Rightarrow \frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}\) (đpcm)

Áp dụng công thức trên (cho tất cả các phần)

a) \(\left\{\begin{matrix} \frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}\\ \frac{1}{b}+\frac{1}{c}\geq \frac{4}{b+c}\\ \frac{1}{c}+\frac{1}{a}\geq \frac{4}{a+c}\end{matrix}\right.\) \(\Rightarrow \) cộng theo về, rút gọn: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\)

Ta có đpcm.

b) Áp dụng CT: \(\left\{\begin{matrix} \frac{1}{a+b}+\frac{1}{a+c}\geq \frac{4}{a+b+a+c}=\frac{4}{2a+b+c}\\ \frac{1}{b+c}+\frac{1}{a+c}\geq \frac{4}{a+b+2c}\\ \frac{1}{a+b}+\frac{1}{b+c}\geq \frac{4}{a+2b+c}\end{matrix}\right.\)

Cộng theo vế và rút gọn:

\(\Rightarrow \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq 2\left (\frac{1}{a+2b+c}+\frac{1}{2a+b+c}+\frac{1}{a+b+2c}\right)\)

Ta có đpcm.

c) Áp dụng hai phần a và b:

\(\text{VP}\leq \frac{1}{2}\left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\leq \frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Leftrightarrow \text{VP}\leq \frac{4}{4}=1\) (đpcm)

Dấu bằng xảy ra ở tất cả các phần đều là khi \(a=b=c\)

Lời giải:

a) Áp dụng BĐT Bunhiacopxky:

\(\text{VT}=(\sqrt{a^3}^2+\sqrt{b^3}^2+\sqrt{c^3}^2)\left (\sqrt{\frac{1}{a}}^2+\sqrt{\frac{1}{b}}^2+\sqrt{\frac{1}{c}}^2\right)\geq (\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2})^2\)

\(\Leftrightarrow \text{VT}\geq (a+b+c)^2\) (đpcm)

b)

Khai triển ta có:

\(3(a^3+b^3+c^3)\geq (a^2+b^2+c^2)(a+b+c)\)

\(\Leftrightarrow 2(a^3+b^3+c^3)\geq ab(a+b)+bc(b+c)+ac(a+c)\)

Áp dụng BĐT Cauchy:

\(a^3+a^3+b^3\geq 3\sqrt[3]{a^6b^3}=3a^2b\)

\(b^3+b^3+c^3\geq 3\sqrt[3]{b^6c3}=3b^2c\)

\(c^3+c^3+a^3\geq 3\sqrt[3]{c^6a^3}=3c^2a\)

Cộng theo vế và rút gọn:

\(\Rightarrow a^3+b^3+c^3\geq a^2b+b^2c+c^2a\)

Hoàn toàn tương tự, ta cũng cm được: \(a^3+b^3+c^3+ab^2+bc^2+ca^2\)

Suy ra \(2(a^3+b^3+c^3)\geq ab(a+b)+bc(b+c)+ac(c+a)\)

Do đó ta có đpcm.

Dấu bằng xảy ra khi $a=b=c$

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left (\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\right)(abc+abc+abc)\geq (ab+bc+ac)^2\)

\(\Leftrightarrow \frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\geq \frac{(ab+bc+ac)^2}{3abc}\) $(1)$

Áp dụng BĐT Cauchy:

\(\left\{\begin{matrix} a^2b^2+b^2c^2\geq 2ab^2c\\ a^2b^2+c^2a^2\geq 2a^2bc\\ b^2c^2+c^2a^2\geq 2abc^2\end{matrix}\right.\Rightarrow a^2b^2+b^2c^2+c^2a^2\geq abc(a+b+c)\)

\(\Leftrightarrow (ab+bc+ac)^2\geq 3abc(a+b+c)(2)\)

Từ \((1),(2)\Rightarrow \frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\geq a+b+c\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c\)

b) Ta có:

\(\text{VT}+3=(a+b+c)\left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

Áp dụng BĐT Bunhiacopxky:

\(\left ( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )(a+b+b+c+c+a)\geq (1+1+1)^2=9\)

\(\Rightarrow \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq \frac{9}{2(a+b+c)}\)

\(\Rightarrow \text{VT}+3\geq (a+b+c).\frac{9}{2(a+b+c)}=\frac{9}{2}\Rightarrow \text{VT}\geq \frac{3}{2}\)

Do đó ta có đpcm.

Lời giải:

a)

Áp dụng BĐT Cauchy:

\((a+b)(b+c)(c+a)\geq 2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ac}=8abc\)

Do đó ta có đpcm. Dấu bằng xảy ra khi \(a=b=c\geq 0\)

b) Áp dụng BĐT Cauchy:

\((a+b+c)(a^2+b^2+c^2)\geq 3\sqrt[3]{abc}.3\sqrt[3]{a^2b^2c^2}=9abc\)

Do đó ta có đpcm. Dấu bằng xảy ra khi \(a=b=c\geq 0\)

c) Áp dụng BĐT Cauchy:

\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}\)

\(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}\)

Cộng theo vế:\(\Rightarrow 3\geq 3\frac{1+\sqrt[3]{abc}}{\sqrt[3]{(a+1)(b+1)(c+1)}}\)

\(\Leftrightarrow (a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3\)

Dấu bằng xảy ra khi $a=b=c$

Ace Legona Akai Haruma Giúp em với .

1) \(P=\left(a+2b+3c\right)\left(6a+3b+2c\right)\)

\(P=\left[a+2b+3\left(1-a-b\right)\right]+\left[6a+3b+2\left(1-a-b\right)\right]=\left(3-2a-b\right)\left(2+4a+b\right)=2\left(3a-2b-b\right)\left(1+2a+\dfrac{b}{2}\right)\)

Lợi dụng AM-GM, ta có:

\(P\le2\left(\dfrac{3-2a-b+1+2a+\dfrac{b}{2}}{2}\right)^2=2.\left(\dfrac{4-\dfrac{b}{2}}{2}\right)^2=8\)

Max_P=8 khi \(a=c=\dfrac{1}{2};b=0\)

2) Sử dụng AM-GM tìm được Max=80 khi b=0;a=2c=2

\(x^2+y^2+z^2=xy+yz+zx\) và \(x^{2009}+y^{2009}+z^{2009}=3^{2010}\)

Ta có:

\(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

Dấu " = " xảy ra :

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\) \(\Rightarrow x=y=z\)

Thay \(x=y=z\) vào \(x^{2009}+y^{2009}+z^{2009}=3^{2009}\) ta được:

\(3x^{2009}=3x^{2010}\)

\(\Rightarrow x^{2009}=3^{2009}\)

\(\Rightarrow x=3\)

\(\Rightarrow y=z=x=3\)

Vậy \(\left(x;y;z\right)=\left(3;3;3\right)\)

Thiếu đề chăng.?

\(S=\dfrac{\sqrt[3]{\left(a-2\right)\left(b-3\right)}}{a+b}\)

\(\Rightarrow S.\sqrt[3]{5}=\dfrac{\sqrt[3]{\left(a-2\right)\left(b-3\right).5}}{a+b}\)

\(\le\dfrac{\dfrac{\left(a-2\right)+\left(b-3\right)+5}{3}}{a+b}=\dfrac{\dfrac{a+b}{3}}{a+b}=\dfrac{1}{3}\)

\(\Rightarrow S\le\dfrac{1}{3}:\sqrt[3]{5}=\dfrac{1}{3\sqrt[3]{5}}\)

Đẳng thức xảy ra \(\Leftrightarrow a-2=b-3=5\Leftrightarrow\left\{{}\begin{matrix}a=7\\b=8\end{matrix}\right.\)

\(P=\dfrac{\sqrt{a-2}}{a}+\dfrac{\sqrt[3]{b-3}}{b}+\dfrac{\sqrt[4]{c-6}}{c}\)

\(=\dfrac{\sqrt{\left(a-2\right).2}}{a\sqrt{2}}+\dfrac{\sqrt[3]{\left(b-3\right).\dfrac{3}{2}.\dfrac{3}{2}}}{b\sqrt[3]{\dfrac{9}{4}}}+\dfrac{\sqrt[4]{\left(c-6\right).2.2.2}}{c\sqrt[3]{8}}\)

\(\le\dfrac{a-2+2}{2a\sqrt{2}}+\dfrac{b-3+\dfrac{3}{2}+\dfrac{3}{2}}{3b\sqrt[3]{\dfrac{9}{4}}}+\dfrac{c-6+2+2+2}{4c\sqrt[4]{8}}\)

\(=\dfrac{a}{2a\sqrt{2}}+\dfrac{b}{3b\sqrt[3]{\dfrac{9}{4}}}+\dfrac{c}{4c\sqrt[4]{8}}=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt[3]{\dfrac{9}{4}}}+\dfrac{1}{4\sqrt[4]{8}}\)

Vậy \(P_{max}=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt[3]{\dfrac{9}{4}}}+\dfrac{1}{4\sqrt[4]{8}}\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-2=2\\b-3=\dfrac{3}{2}\\c-6=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=\dfrac{9}{2}\\c=8\end{matrix}\right.\)

\(P=\dfrac{bc\sqrt{a-2}+ac\sqrt[3]{b-3}+ab\sqrt[4]{c-6}}{abc}\)

\(=\dfrac{\sqrt{a-2}}{a}+\dfrac{\sqrt[3]{b-3}}{b}+\dfrac{\sqrt[4]{c-6}}{c}\)

Áp dụng BĐT AM-GM ta có:

\(=\dfrac{\sqrt{2\left(a-2\right)}}{\sqrt{2}a}+\dfrac{\sqrt[3]{2\left(b-3\right)}}{\sqrt[3]{2}b}+\dfrac{\sqrt[4]{2\left(c-6\right)}}{\sqrt[4]{2}c}\)

\(\le\dfrac{\dfrac{2+a-2}{2}}{\sqrt{2}a}+\dfrac{\dfrac{2+b-3+1}{3}}{\sqrt[3]{2}b}+\dfrac{\dfrac{2+c-6+1+1+1+1}{4}}{\sqrt[4]{2}c}\)

\(=\dfrac{\dfrac{a}{2}}{\sqrt{2}a}+\dfrac{\dfrac{b}{3}}{\sqrt[3]{2}b}+\dfrac{\dfrac{c}{4}}{\sqrt[4]{2}c}=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt[3]{2}}+\dfrac{1}{4\sqrt[4]{2}}\)