Chọn D

Với mọi số thực  x; y; z ta có: \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\) ( tự chứng minh xem; có thể áp dụng )

Ta có: \(S^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)

\(\le3\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]=6\left(a+b+c\right)=6\)

=> \(S\le\sqrt{6}\)

Dấu "=" xảy ra <=> a = b = c =1/3

Vậy max S = \(\sqrt{6}\) tại a = b = c = 1/3.

đây nhé bạn

Chọn đáp án B

Sử dụng bất đẳng thức Cauchy – Schwarz dạng phân thức ta có

Cách 2: Ghép cặp và dùng BĐT Cauchy. Cụ thể 

 Giả thiết trở thành 

Ta đi tìm GTLN của 

Sử dụng bất đẳng thức Cauchy – Schwarz dạng phân thức ta có

Suy ra 

Chọn B.

Cách 2. Ghép cặp và dùng BĐT Cauchy. Cụ thể 

Bài 1: Ta có:

\(M=\frac{ad}{abcd+abd+ad+d}+\frac{bad}{bcd.ad+bc.ad+bad+ad}+\frac{c.abd}{cda.abd+cd.abd+cabd+abd}+\frac{d}{dab+da+d+1}\)

\(=\frac{ad}{1+abd+ad+d}+\frac{bad}{d+1+bad+ad}+\frac{1}{ad+d+1+abd}+\frac{d}{dab+da+d+1}\)

$=\frac{ad+abd+1+d}{ad+abd+1+d}=1$

Bài 2:

Vì $a,b,c,d\in [0;1]$ nên

\(N\leq \frac{a}{abcd+1}+\frac{b}{abcd+1}+\frac{c}{abcd+1}+\frac{d}{abcd+1}=\frac{a+b+c+d}{abcd+1}\)

Ta cũng có:
$(a-1)(b-1)\geq 0\Rightarrow a+b\leq ab+1$

Tương tự:

$c+d\leq cd+1$

$(ab-1)(cd-1)\geq 0\Rightarrow ab+cd\leq abcd+1$

Cộng 3 BĐT trên lại và thu gọn thì $a+b+c+d\leq abcd+3$

$\Rightarrow N\leq \frac{abcd+3}{abcd+1}=\frac{3(abcd+1)-2abcd}{abcd+1}$

$=3-\frac{2abcd}{abcd+1}\leq 3$

Vậy $N_{\max}=3$

3.

Hình vẽ:

Lời giải:

a) △AMC và △BNC có: \(\widehat{AMC}=\widehat{BNC}=90^0;\widehat{ACB}\) là góc chung.

\(\Rightarrow\)△AMC∼△BNC (g-g).

\(\Rightarrow\dfrac{AC}{BC}=\dfrac{CM}{CN}\Rightarrow AC.CN=BC.CM\left(1\right)\)

b) △AMB và △CPB có: \(\widehat{AMB}=\widehat{CPB}=90^0;\widehat{ABC}\) là góc chung.

\(\Rightarrow\)△AMB∼△CPB (g-g)

\(\Rightarrow\dfrac{AB}{CB}=\dfrac{BM}{BP}\Rightarrow AB.BP=BC.BM\left(2\right)\)

Từ (1) và (2) suy ra:

\(AC.CN+AB.BP=BC.CM+BC.BM=BC.\left(CM+BM\right)=BC.BC=BC^2\left(đpcm\right)\)b) Gọi \(M_0\) là trung điểm BC, giả sử \(AB< AC\).

\(\widehat{HBM}=90^0-\widehat{BHM}=90^0-\widehat{AHN}=\widehat{CAM}\)

△HBM và △CAM có: \(\widehat{HBM}=\widehat{CAM};\widehat{HMB}=\widehat{CMA}=90^0\)

\(\Rightarrow\)△HBM∼△CAM (g-g) 

\(\Rightarrow\dfrac{MH}{CM}=\dfrac{BM}{MA}\Rightarrow MH.MA=BM.CM\)

Ta có: \(BM.CM=\left(BM_0-MM_0\right)\left(CM_0+MM_0\right)=\left(BM_0-MM_0\right)\left(BM_0+MM_0\right)=BM_0^2-MM_0^2\le BM_0^2=\dfrac{BC^2}{4}\)

\(\Rightarrow MH.MA\le\dfrac{BC^2}{4}\).

Vì \(BC\) không đổi nên: \(max\left(MH.MA\right)=\dfrac{BC^2}{4}\), đạt được khi △ABC cân tại A hay A nằm trên đường trung trực của BC.

c) Sửa đề: \(S_1.S_2.S_3\le\dfrac{1}{64}.S^3\)

△AMC∼△BNC \(\Rightarrow\dfrac{AC}{BC}=\dfrac{MC}{NC}\Rightarrow\dfrac{AC}{MC}=\dfrac{BC}{NC}\)

△ABC và △MNC có: \(\dfrac{AC}{MC}=\dfrac{BC}{NC};\widehat{ACB}\) là góc chung.

\(\Rightarrow\)△ABC∼△MNC (c-g-c)

\(\Rightarrow\dfrac{S_{MNC}}{S_{ABC}}=\dfrac{S_1}{S}=\dfrac{MC}{AC}.\dfrac{NC}{BC}\left(1\right)\)

Tương tự: 

△ABC∼△MBP \(\Rightarrow\dfrac{S_{MBP}}{S_{ABC}}=\dfrac{S_2}{S}=\dfrac{MB}{AB}.\dfrac{BP}{BC}\left(2\right)\)

△ABC∼△ANP \(\Rightarrow\dfrac{S_{ANP}}{S_{ABC}}=\dfrac{S_3}{S}=\dfrac{AN}{AB}.\dfrac{AP}{AC}\left(3\right)\)

Từ (1), (2), (3) suy ra:

\(\dfrac{S_1}{S}.\dfrac{S_2}{S}.\dfrac{S_3}{S}=\left(\dfrac{MC}{AC}.\dfrac{NC}{BC}\right).\left(\dfrac{MB}{AB}.\dfrac{BP}{BC}\right).\left(\dfrac{AN}{AB}.\dfrac{AP}{AC}\right)\) 

\(\Rightarrow\dfrac{S_1}{S}.\dfrac{S_2}{S}.\dfrac{S_3}{S}=\left(\dfrac{MC.MB}{AC.AB}\right).\left(\dfrac{BP.AP}{AC.BC}\right).\left(\dfrac{AN.CN}{AB.BC}\right)\) (*)

Áp dụng câu b) ta có:

\(\left\{{}\begin{matrix}BM.CM\le\dfrac{1}{4}BC^2\\AP.BP\le\dfrac{1}{4}AB^2\\AN.CN\le\dfrac{1}{4}AC^2\end{matrix}\right.\)

Từ (*) suy ra:

\(\dfrac{S_1}{S}.\dfrac{S_2}{S}.\dfrac{S_3}{S}\le\left(\dfrac{\dfrac{1}{4}BC^2}{AC.AB}\right).\left(\dfrac{\dfrac{1}{4}AC^2}{AC.BC}\right).\left(\dfrac{\dfrac{1}{4}AB^2}{AB.BC}\right)=\dfrac{1}{64}\)

\(\Rightarrow S_1.S_2.S_3\le\dfrac{1}{64}.S^3\)

Dấu "=" xảy ra khi △ABC đều.

c1:áp dụng bđt AM-GM:

\(a+b\ge2\sqrt{ab}\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2=1008^2\)

=> đáp án A

c2: tương tự c1 . đáp án b

3.

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ab}}=2\)

Đáp án A

4.

\(a^2-a+1=\left(a-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) ;\(\forall a\)

Đáp án A

Áp dụng cô si ,ta có

\(a^2+b^2\ge2ab\)

\(c^2+b^2\ge2bc\)

\(a^2+c^2\ge2ac\)

\(\Rightarrow2a^2+2b^2+2c^2\ge2ab+2ac+2bc\)

\(a^2+b^2+c^2\ge ab+ac+bc\)

\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc\ge3ab+3ac+3bc\)

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)

\(\Rightarrow200^2\ge3\left(ab+bc+ac\right)\)

\(\Rightarrow\frac{40000}{3}\ge ab+bc+ac\)

Dấu = xảy ra khi a=b=c=200/3

                      Bài làm :

Ta có :

\(\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\frac{4}{a+b}\le\frac{1}{a}+\frac{1}{b}\)

\(\Leftrightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\left(1\right)\)

Dấu "=" xảy ra khi : a=b

Chứng minh tương tự như trên ; ta có :

\(\hept{\begin{cases}\frac{1}{b+c}\text{≤}\frac{1}{4}\left(\frac{1}{b}+\frac{1}{c}\right)\left(2\right)\\\frac{1}{c+a}\text{≤}\frac{1}{4}\left(\frac{1}{c}+\frac{1}{a}\right)\left(3\right)\end{cases}}\)

Cộng vế với vế của (1) ; (2) ; (3) ; ta được :

\(A\text{≤}\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\text{=}\frac{3}{2}\)

Dấu "=" xảy ra khi ;

\(\hept{\begin{cases}a=b=c\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\end{cases}}\Leftrightarrow a=b=c=1\)

Vậy Max (A) = 3/2 khi a=b=c=1

quản lí tên kiểu j z

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