Em tham khảo ở đây:

Cho a,b,c > 0 và ab + bc + ac = 1. Chứng minh rằng :\(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^... - Hoc24

Câu 2a

\(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)

\(\Leftrightarrow a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=\left(a^2+b^2\right)c^2+d^2\left(a^2+b^2\right)\)

\(\Leftrightarrow a^2c^2+b^2d^2+a^2d^2+b^2c^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)

\(\Leftrightarrow a^2c^2+b^2d^2+a^2d^2+b^2c^2-\left(a^2c^2+b^2d^2+a^2d^2+b^2c^2\right)=0\)

\(\Leftrightarrow0=0\)( đpcm )

Câu 2b

\(\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)

\(\Leftrightarrow a^2c^2+2abcd+b^2d^2\le\left(a^2+b^2\right)c^2+d^2\left(a^2+b^2\right)\)

\(\Leftrightarrow a^2c^2+2abcd+b^2d^2\le a^2c^2+b^2c^2+a^2d^2+b^2d^2\)

\(\Leftrightarrow2abcd\le b^2c^2+a^2d^2\)

\(\Leftrightarrow0\le b^2c^2-2abcd+a^2d^2\)

\(\Leftrightarrow0\le\left(bc-ad\right)^2\)( đpcm )

Câu 4a

\(\frac{a+b}{2}\ge\sqrt{ab}\)

\(\Leftrightarrow\left(\frac{a+b}{2}\right)^2\ge ab\)

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

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

\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)( đpcm )

Câu 4c 

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow3a+5b\ge2\sqrt{3a.5b}=2\sqrt{15ab}\)

\(\Rightarrow12\ge2\sqrt{15ab}\)

\(\Rightarrow6\ge\sqrt{15ab}\)

\(\Rightarrow6^2\ge15ab\)

\(\Rightarrow36\ge15ab\)

\(\Rightarrow ab\le\frac{12}{5}\)

\(\Leftrightarrow P\le\frac{12}{5}\)

Vậy GTLN  của \(P=\frac{12}{5}\)

1, Giả sử \(\sqrt{7}\)là số hữu tỷ, khi đó ta có: \(\sqrt{7}=\frac{m}{n},\left(m,n\right)=1\)

Vậy: \(m^2=7n^2\Rightarrow m^2⋮7\Rightarrow m⋮7\Rightarrow m=7k,k\in Z\)

\(\Rightarrow m^2=49k^2\Rightarrow n^2⋮7\Rightarrow n⋮7\)

Vậy (m,n)=7 vô lý do (m,n)=1

Vậy \(\sqrt{7}\)là số vô tỷ

3, Có: \(x^2+y^2\ge2xy\Leftrightarrow2\left(x^2+y^2\right)\ge x^2+2xy+y^2=\left(x+y\right)^2=4\Rightarrow x^2+y^2\ge2\)

VaayjjMin S=2 <=>  x=y=1

4c, Áp dụng BĐT Cauchy: \(3a+5b\ge2\sqrt{3.5ab}\Leftrightarrow ab\le\frac{\left(3a+5b\right)^2}{4.3.5}=\frac{12^2}{4.3.5}=\frac{12}{5}\)

Dấu =xảy ra: \(\hept{\begin{cases}3a=5b\\3a+5b=12\end{cases}\Leftrightarrow}\hept{\begin{cases}a=2\\b=\frac{6}{5}\end{cases}}\)

5, \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=1-3ab\)

Lại có: \(a^2+b^2\ge2ab\Leftrightarrow\left(a+b^2\right)\ge4ab\Leftrightarrow ab\le\frac{1}{4}\)

Vậy \(P\ge1-3.\frac{1}{4}=\frac{1}{4}\)

Vậy Min P=1/4 <=> a=b=1/2

6, Ta có: \(a^3+1+1\ge3\sqrt[3]{a^3.1.1}=3a\)
Tương tự cho b, cộng vế theo vế ta được: \(a^3+b^3+4\ge3\left(a+b\right)\Leftrightarrow a+b\le\frac{a^3+b^3+4}{3}=\frac{2+4}{3}=2\)

Vậy Max N=2 ,+. a=b=1

7, Ta sẽ CM: \(a^3+b^3\ge ab\left(a+b\right)\)

Thật vậy, Xét: \(a^3+b^3-ab\left(a+b\right)=a^2\left(a-b\right)-b^2\left(a-b\right)=\left(a-b\right)\left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)^2\ge0,\forall a,b>0\)Vậy Ta có: \(a^3+b^3+abc\ge ab\left(a+b\right)+abc=ab\left(a+b+c\right)\left(ĐPCM\right)\)

8, Có: I a+b I> Ia-b I>0 <=> \(\left(a+b\right)^2>\left(a-b\right)^2\Leftrightarrow\left(a+b\right)^2-\left(a-b\right)^2>0\Leftrightarrow4ab>0\Leftrightarrow ab>0\)

Vậy Với ab>0 thì ta có BĐT trên

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.