Vi a + b + c = 1 nên bt tương đương với \(P=abc\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)

Ta có : \(P=abc\left(a+b+c\right)\left(a^2+b^2+c^2\right)\le\frac{1}{3}\left(ab+bc+ca\right)^2\left(a^2+b^2+c^2\right)\)( 1 ) 

Mặt khác :\(\left(ab+bc+ca\right)^2\left(a^2+b^2+c^2\right)\le\left(\frac{\left(a+b+c\right)^2}{3}\right)^3=\frac{1}{27}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow P\le\frac{1}{3}.\frac{1}{27}=\frac{1}{81}\)

Dấu "=" xảy ra <=> a = b = c = 1/3

Vậy maxP = 1/81 <=> a = b = c = 1/3

\(P=\dfrac{1}{abc}+\dfrac{1}{a^2+b^2+c^2}=\dfrac{a+b+c}{abc}+\dfrac{1}{a^2+b^2+c^2}\)

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

\(\)\(\left\{{}\begin{matrix}a+b+c=1\\\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{9}{ab+bc+ac}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow P\ge\dfrac{9}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)

\(=\dfrac{1}{2\left(ab+bc+ac\right)}+\dfrac{1}{a^2+b^2+c^2}+\dfrac{17}{2\left(ab+bc+ac\right)}\)

\(\Rightarrow P\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{17}{2\left(ab+bc+ac\right)}\)

\(\Rightarrow P\ge9+\dfrac{17}{2\left(ab+bc+ac\right)}\)

mà \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\)

\(\Rightarrow P\ge9+\dfrac{17}{2.\dfrac{1}{3}}=9+\dfrac{17.3}{2}=\dfrac{18+17.3}{2}=\dfrac{69}{2}\)

\(\Rightarrow Min\left(P\right)=\dfrac{69}{2}\)

Ta có a² + 1 \(\ge\)2a 

Khi đó \(\frac{1}{a^2+ab-a+5}=\frac{1}{a^2+1+ab-a+4}\le\frac{1}{2a+ab-a+4}=\frac{1}{ab+a+4}\)

Tương tự ta được \(\frac{1}{b^2+bc-b+5}\le\frac{1}{bc+b+4};\frac{1}{c^2+ac-c+5}\le\frac{1}{ac+c+4}\)

Cộng vế với vế => A \(\le\frac{1}{ab+a+4}+\frac{1}{bc+b+4}+\frac{1}{ca+c+4}\)

=> 4A \(\le\frac{4}{ab+a+1+3}+\frac{4}{bc+b+1+3}+\frac{4}{ca+c+1+3}\)

\(\le\frac{1}{ab+a+1}+\frac{1}{3}+\frac{1}{bc+b+1}+\frac{1}{3}+\frac{1}{ac+a+1}+\frac{1}{3}\)

\(=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+a+1}+1\)

\(=\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{a^2bc+abc+ab}+1\)

\(=\frac{1}{ab+a+1}+\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+1=\frac{ab+a+1}{ab+a+1}+1=1+1=2\)

=> \(A\le\frac{1}{2}\)(Dấu "=" xảy ra <=> a = b = c = 1)

cho mik hỏi tí là làm sao ra được \(\frac{4}{ab+a+1+3}\le\frac{1}{ab+a+1}+\frac{1}{3}\) vậy ạ?

Dự đoán điểm rơi a = b = c = 1 

Ta có : \(\frac{1}{ab+a+1}+\frac{1}{3}\ge\frac{\left(1+1\right)^2}{ab+a+1+3}\)(BĐT Schwarz) 

\(=\frac{4}{a+b+c+4}\) (đpcm) 

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.

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