Bài 1:

Áp dụng BĐT Bunhiacopxky ta có:

$(a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2$

$\Leftrightarrow 3(a^2+b^2+c^2)\geq 1$

$\Leftrightarrow a^2+b^2+c^2\geq \frac{1}{3}$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

Bài 2: 

Áp dụng BĐT Bunhiacopxky:

$(a^2+4b^2+9c^2)(1+\frac{1}{4}+\frac{1}{9})\geq (a+b+c)^2$

$\Leftrightarrow 2015.\frac{49}{36}\geq (a+b+c)^2$

$\Leftrightarrow \frac{98735}{36}\geq (a+b+c)^2$

$\Rightarrow a+b+c\leq \frac{7\sqrt{2015}}{6}$ chứ không phải $\frac{\sqrt{14}}{6}$ :''>>

Bài 3:

Áp dụng BĐT Bunhiacopxky:

$2=(a^2+b^2)(1+1)\geq (a+b)^2\Rightarrow a+b\leq \sqrt{2}$

$(a\sqrt{1+a}+b\sqrt{1+b})^2\leq (a^2+b^2)(1+a+1+b)$

$=2+a+b\leq 2+\sqrt{2}$

$\Rightarrow a\sqrt{1+a}+b\sqrt{1+b}\leq \sqrt{2+\sqrt{2}}$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=\frac{1}{\sqrt{2}}$

thử bài bất :D 

Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)

Hoàn toàn tương tự: 

\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)

\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)

Cộng (*),(**),(***) vế theo vế ta được:

\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)

Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )

Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi a=b=c=1

1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D 

Ta có:

Dấu “=” xảy ra khi và chỉ khi

Vậy số bộ a,b,c thỏa mãn điều kiện đã cho là 1.

Chọn B.

1.

Ta sẽ chứng minh BĐT sau: \(\dfrac{1}{a^2+b^2}+\dfrac{1}{b^2+c^2}+\dfrac{1}{c^2+a^2}\ge\dfrac{10}{\left(a+b+c\right)^2}\)

Do vai trò a;b;c như nhau, ko mất tính tổng quát, giả sử \(c=min\left\{a;b;c\right\}\)

Đặt \(\left\{{}\begin{matrix}x=a+\dfrac{c}{2}\\y=b+\dfrac{c}{2}\end{matrix}\right.\) \(\Rightarrow x+y=a+b+c\)

Đồng thời \(b^2+c^2=\left(b+\dfrac{c}{2}\right)^2+\dfrac{c\left(3c-4b\right)}{4}\le\left(b+\dfrac{c}{2}\right)^2=y^2\)

Tương tự: \(a^2+c^2\le x^2\) ; \(a^2+b^2\le x^2+y^2\)

Do đó: \(A\ge\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2+y^2}\)

Nên ta chỉ cần chứng minh: \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2+y^2}\ge\dfrac{10}{\left(x+y\right)^2}\)

Mà \(\dfrac{1}{\left(x+y\right)^2}\le\dfrac{1}{4xy}\) nên ta chỉ cần chứng minh:

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2+y^2}\ge\dfrac{5}{2xy}\)

\(\Leftrightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{2}{xy}+\dfrac{1}{x^2+y^2}-\dfrac{1}{2xy}\ge0\)

\(\Leftrightarrow\dfrac{\left(x-y\right)^2}{x^2y^2}-\dfrac{\left(x-y\right)^2}{2xy\left(x^2+y^2\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(x-y\right)^2\left(2x^2+2y^2-xy\right)}{2x^2y^2}\ge0\) (luôn đúng)

Vậy \(A\ge\dfrac{10}{\left(a+b+c\right)^2}\ge\dfrac{10}{3^2}=\dfrac{10}{9}\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{3}{2};\dfrac{3}{2};0\right)\) và các hoán vị của chúng

2.

Ta có: \(B=\dfrac{ab+1-1}{1+ab}+\dfrac{bc+1-1}{1+bc}+\dfrac{ca+1-1}{1+ca}\)

\(B=3-\left(\dfrac{1}{1+ab}+\dfrac{1}{1+ca}+\dfrac{1}{1+ab}\right)\)

Đặt \(C=\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\)

Ta có: \(C\ge\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{27}{13}\)

\(\Rightarrow B\le3-\dfrac{27}{13}=\dfrac{12}{13}\)

\(B_{max}=\dfrac{12}{13}\) khi \(a=b=c=\dfrac{2}{3}\)

Do \(a;b;c\in\left[0;1\right]\)

\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\)\(\Leftrightarrow ab+1\ge a+b\)

\(\Leftrightarrow ab+c+1\ge a+b+c=2\)

\(\Rightarrow abc+ab+c+1\ge ab+c+1\ge2\)

\(\Rightarrow\left(c+1\right)\left(ab+1\right)\ge2\)

\(\Rightarrow\dfrac{1}{ab+1}\le\dfrac{c+1}{2}\)

Hoàn toàn tương tự, ta có: 

\(\dfrac{1}{bc+1}\le\dfrac{a+1}{2}\) ; \(\dfrac{1}{ca+1}\le\dfrac{b+1}{2}\)

Cộng vế: \(C\le\dfrac{a+b+c+3}{2}=\dfrac{5}{2}\)

\(\Rightarrow B\ge3-\dfrac{5}{2}=\dfrac{1}{2}\)

\(B_{min}=\dfrac{1}{2}\) khi \(\left(a;b;c\right)=\left(0;1;1\right)\) và các hoán vị của chúng

Bài 3:

\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)

\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)

-Tham khảo:

-Tham khảo:

Áp dụng BĐT cô si với hai số không âm, Ta có: 

\(\left(a+b+c\right)^2=1\ge4a\left(b+c\right)\)

\(\Leftrightarrow b+c\ge4a\left(b+c\right)^2\)

Mà \(\left(b+c\right)^2\ge4bc\forall b,c\ge0\)

\(\Rightarrow b+c\ge16abc\)

Dấu "=" xảy ra khi: 

\(\hept{\begin{cases}a+b+c=1\\b=c\\a=b+c\end{cases}}\Rightarrow\hept{\begin{cases}a=\frac{1}{2}\\b=c=\frac{1}{4}\end{cases}}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Leftrightarrow3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge9\)

\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge6\)

Áp dụng BĐT Cô si với 2 số dương ta có: 

\(\frac{a}{b}+\frac{b}{a}\ge2,\frac{b}{c}+\frac{c}{b}\ge2,\frac{c}{a}+\frac{a}{c}\ge2\)

\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge6\)(đúng) 

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)(do a+b+c=1)

\(a,\dfrac{3}{a+b}=\dfrac{2}{b+c}=\dfrac{1}{c+a}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{b+c}{2}=\dfrac{c+a}{1}=\dfrac{2\left(a+b+c\right)}{6}=\dfrac{a+b+c}{3}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{a+b+c}{3}\\ \Rightarrow3\left(a+b+c\right)=3\left(a+b\right)\\ \Rightarrow3\left(a+b\right)+3c=3\left(a+b\right)\\ \Rightarrow3c=0\\ \Rightarrow c=0\)

Vậy \(P=\dfrac{a+b-2019c}{a+b+2018c}=\dfrac{a+b}{a+b}=1\)

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.

1.

\(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^2=0\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)

Ta có:

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

\(=\dfrac{a^2+4b^2+4ab+b^2+4c^2+4bc+c^2+4a^2+4ca}{a^2+4b^2-4ab+b^2+4c^2-4bc+c^2+4a^2-4ca}\)

\(=\dfrac{5\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)}{5\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)}\)

\(=\dfrac{-10\left(ab+bc+ca\right)+4\left(ab+bc+ca\right)}{-10\left(ab+bc+ca\right)-4\left(ab+bc+ca\right)}\)

\(=\dfrac{-6}{-14}=\dfrac{3}{7}\)

b.

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3abc\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)

\(\Rightarrow\dfrac{ab+2bc+3ca}{3a^2+4b^2+5c^2}=\dfrac{a^2+2a^2+3a^2}{3a^2+4a^2+5a^2}=\dfrac{6}{12}=\dfrac{1}{2}\)