Ta co:

\(M=\frac{9}{1-2\left(ab+bc+ca\right)}+\frac{2}{abc}=\frac{9}{\left(a+b+c\right)^2-2\left(ab+bc+ca\right)}+\frac{2}{abc}=\frac{9}{a^2+b^2+c^2}+\frac{2}{abc}\)

Ta lai co:

\(a+b+c=1\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{1}{abc}\)

\(\Rightarrow M=\frac{9}{\Sigma_{cyc}a^2}+\Sigma_{cyc}\frac{2}{ab}\ge\frac{9}{\Sigma_{cyc}a^2}+\frac{18}{\Sigma_{cyc}ab}\left(1\right)\)

\(VT_{\left(1\right)}=\frac{9}{\Sigma_{cyc}a^2}+\frac{1}{\Sigma_{cyc}ab}+\frac{1}{\Sigma_{cyc}ab}+\frac{16}{\Sigma_{cyc}ab}\ge\frac{\left(3+1+1\right)^2}{\Sigma_{cyc}a^2+2\Sigma_{cyc}ab}+\frac{16}{\frac{\left(\Sigma_{cyc}a\right)^2}{3}}=\text{ }\frac{25}{\left(\Sigma_{cyc}a\right)^2}+48=\text{ }73\)

Dau '=' xay ra khi \(\text{ }a=b=c=\frac{1}{3}\)

@my-friend 

\(M\ge\frac{9}{a^2+b^2+c^2}+\frac{36}{2\left(ab+bc+ca\right)}\ge\frac{\left(3+6\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=81\)

Dấu "=" xảy ra ra khi \(\hept{\begin{cases}\frac{3}{a^2+b^2+c^2}=\frac{6}{2\left(ab+bc+ca\right)}\\a+b+c=1\end{cases}}\Leftrightarrow a=b=c=\frac{1}{3}\)

Trước hết dễ có: \(\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\ge abc\)

\(\Rightarrow abc\le\frac{ab+bc+ca}{9}=t\) (với \(0< t=\frac{ab+bc+ca}{9}\le\frac{\left(a+b+c\right)^2}{27}=\frac{1}{27}\))

Do đó \(M\ge\frac{9}{1-18t}+\frac{2}{t}=\frac{2\left(27t-1\right)^2}{t\left(1-18t\right)}+81\ge81\forall0< t\le\frac{1}{27}\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)

\(P=\frac{a^3}{\left(a+1\right)\left(b+1\right)}+\frac{b^3}{\left(b+1\right)\left(c+1\right)}+\frac{c^3}{\left(c+1\right)\left(a+1\right)}-1\)

ôi trá hình :VVV

\(P=\frac{a^3}{\left(a+1\right).\left(b+1\right)}+\frac{b^3}{\left(b+1\right).\left(c+1\right)}+\frac{c^3}{\left(c+1\right).\left(a+1\right)}\)

Ko biết đúng hay không!

Mới lớp 6 , mà tôi nghĩ Lầy Văn Lội đúng đấy!

Bài toán số 41 có 2 cách làm, tôi làm cách thứ 2

Đặt \(Q=\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{x+z}}+\sqrt{\frac{z}{x+y}}\)\(\Rightarrow Q^2=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}+2\left(\sqrt{\frac{xy}{\left(y+z\right)\left(x+z\right)}}+\sqrt{\frac{yz}{\left(x+z\right)\left(y+z\right)}}+\sqrt{\frac{xz}{\left(x+y\right)\left(y+z\right)}}\right)\)ta thấy rằng \(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{1}{4}\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)\left(xy+yz+zx\right)\)

\(=\frac{x^2+y^2+z^2}{4}+\frac{xyz}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge\frac{x^2+y^2+z^2}{4}\)

Áp dụng bất đẳng thức AM-GM ta có \(\sqrt{\frac{yx}{\left(z+x\right)\left(x+y\right)}}\ge\frac{2yx}{2\sqrt{\left(xy+yz\right)\left(yz+yx\right)}}\ge\frac{2xy}{2xy+yz+xz}\ge\frac{2xy}{2\left(xy+yz+zx\right)}=\frac{xy}{xy+yz+zx}\)

Tương tự ta có \(\hept{\begin{cases}\sqrt{\frac{yz}{\left(z+x\right)\left(z+y\right)}}\ge\frac{yz}{xy+yz+zx}\\\sqrt{\frac{xz}{\left(x+y\right)\left(y+z\right)}}\ge\frac{xz}{xy+yz+zx}\end{cases}}\)

\(\Rightarrow\sqrt{\frac{xy}{\left(y+z\right)\left(z+x\right)}}+\sqrt{\frac{yz}{\left(z+x\right)\left(x+y\right)}}+\sqrt{\frac{zx}{\left(x+y\right)\left(y+z\right)}}\ge1\)nên \(Q\ge\sqrt{\frac{x^2+y^2+z^2}{4}+2}\)

\(\Rightarrow Q\ge\sqrt{\frac{x^2+y^2+z^2}{2}+4}+\frac{4}{\sqrt{x^2+y^2+z^2}}\)

Đặt \(t=\sqrt{x^2+y^2+z^2}\Rightarrow t\ge\sqrt{xy+yz+zx}=2\)

Xét hàm số g(t)=\(\sqrt{\frac{t^2}{2}+4}+\frac{4}{t}\left(t\ge2\right)\)khi đó ta có 

\(g'\left(t\right)=\frac{t}{2\sqrt{\frac{t^2}{2}+4}}-\frac{4}{t^2};g'\left(t\right)=0\Leftrightarrow t^6-32t^2-256=0\Leftrightarrow t=2\sqrt{2}\)

Lập bảng biến thiên ta có min[2;\(+\infty\)) \(g\left(t\right)=g\left(2\sqrt{2}\right)=3\sqrt{2}\)

Hay minS=\(3\sqrt{2}\)<=> a=c=1; b=2

Đặt a=xc; b=cy (x;y >=1)

\(\sqrt{x-1}+\sqrt{y-1}=\sqrt{xy}\Leftrightarrow x+y-2+2\sqrt{\left(x-1\right)\left(y-1\right)}=xy\)

\(\Leftrightarrow xy-x-y+1-2\sqrt{\left(x-1\right)\left(y-1\right)}+1=0\)

\(\Leftrightarrow\left(\sqrt{\left(x-1\right)\left(y-1\right)}-1\right)^2=0\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(y-1\right)}=1\Leftrightarrow xy=x+y\ge2\sqrt{xy}\Rightarrow xy\ge4\)

Biểu thức P được viết lại như sau

\(P=\frac{x}{y+1}+\frac{y}{x+1}+\frac{1}{x+y}+\frac{1}{x^2+y^2}=\frac{x^2}{xy+x}+\frac{y^2}{xy+y}+\frac{1}{x^2+y^2}+\frac{1}{\left(x+y\right)^2-2xy}\)

\(P\ge\frac{\left(x+y\right)^2}{2xy+x+y}+\frac{1}{x+y}+\frac{1}{\left(x+y\right)^2-2xy}=\frac{xy}{3}+\frac{1}{xy}+\frac{1}{x^2y^2-2xy}=\frac{x^3y^3-2x^2y^2+3xy-3}{3\left(x^2y^2-2xy\right)}\)

Đặt t=xy với t>=4

Xét hàm số \(f\left(t\right)=\frac{t^3-2t^2+3t-3}{t^2-2t}\left(t\ge4\right)\)

Ta có \(f'\left(t\right)=\frac{t^4-4t^3+t^2+6t-6}{\left(t^2-2t\right)^2}=\frac{t^3\left(t-4\right)+6\left(t-4\right)+18}{\left(t^2-2t\right)^2}>0\forall t\ge4\)

Lập bảng biến thiên ta có \(minf\left(t\right)=f\left(4\right)=\frac{41}{8}\)

Vậy \(minP=\frac{41}{24}\)khi x=y=z=2 hay a=b=2c

Bài 1 quan trong là đoán dấu đẳng thức.

1/  Có: \(36=\left(3+2+1\right)\left(a^2+b^2+c^2\right)\ge\left(\sqrt{3}a+\sqrt{2}b+c\right)^2\)

\(\therefore\sqrt{3}a+\sqrt{2}b+c\le6\)

\(\frac{1}{3}\left(\frac{a}{bc}+\frac{3b}{2ca}\right)+\frac{3}{2}\left(\frac{b}{ca}+\frac{2c}{ab}\right)+2\left(\frac{c}{ab}+\frac{a}{3bc}\right)\)

\(\ge\frac{\sqrt{6}}{3c}+\frac{3\sqrt{2}}{a}+\frac{4\sqrt{3}}{3b}\)

\(=\frac{\left(\frac{\sqrt{6}}{3}\right)}{c}+\frac{\left(3\sqrt{6}\right)}{\sqrt{3}a}+\frac{\left(\frac{4\sqrt{6}}{3}\right)}{\sqrt{2}b}\)

\(\ge\frac{\left(\sqrt{\frac{\sqrt{6}}{3}}+\sqrt{3\sqrt{6}}+\sqrt{\frac{4\sqrt{6}}{3}}\right)^2}{\sqrt{3}a+\sqrt{2}b+c}\ge2\sqrt{6}\)

Đẳng thức xảy ra khi \(a=\sqrt{3},b=\sqrt{2},c=1\)

Hiếm hoi thấy anh tth làm bất ko dùng sos

Tại bài này SOS làm biếng thôi em, lằng nhằng lắm

1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)

\(=ac+bc+c^2+ab\)

\(=a\left(b+c\right)+c\left(b+c\right)\)

\(=\left(b+c\right)\left(a+b\right)\)

CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)

\(b+ca=\left(b+c\right)\left(a+b\right)\)

Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)

Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)

CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)

\(\Rightarrow P\le\frac{1}{2}.3\)

\(\Rightarrow P\le\frac{3}{2}\)

Dấu"="xảy ra \(\Leftrightarrow a=b=c\)

Vậy /...

\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)

\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)

Tương tự rồi cộng lại:

\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)

\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)

Dấu "=" xảy ra tại \(a=b=c=1\)

Bài 1 : 

\(P=\sqrt{\frac{ab}{c+ab}}+\sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}\)

\(P=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\frac{bc}{a\left(a+b+c\right)+bc}}\)

\(+\sqrt{\frac{ca}{b\left(a+b+c\right)+ca}}\)

\(P=\sqrt{\frac{ab}{ac+bc+c^2+ab}}+\sqrt{\frac{bc}{a^2+ab+ac+bc}}\)

\(+\sqrt{\frac{ca}{ab+b^2+bc+ca}}\)

\(P=\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\)

Áp dụng bất đẳng thức Cauchy cho 2 bô só thực không âm

\(\Rightarrow\hept{\begin{cases}\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\le\frac{\frac{a}{a+c}+\frac{b}{b+c}}{2}\\\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}\\\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\le\frac{\frac{a}{a+b}+\frac{c}{b+c}}{2}\end{cases}}\)

\(\Rightarrow VT\)

\(\le\frac{\left(\frac{a}{a+c}+\frac{c}{a+c}\right)+\left(\frac{b}{b+c}+\frac{c}{b+c}\right)+\left(\frac{b}{a+b}+\frac{a}{a+b}\right)}{2}\)

\(\Rightarrow VT\le\frac{\frac{a+c}{a+c}+\frac{b+c}{b+c}+\frac{a+b}{a+b}}{2}=\frac{3}{2}\)

\(\Rightarrow P\le\frac{3}{2}\)

Vậy \(P_{max}=\frac{3}{2}\)

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

Chúc bạn học tốt !!!

Bunhia thì phải hoặc tương đương thần chưởng @@
Có lẽ bunhia đấy :vv

Câu này t dùng vi-et giải được. Nhưng để mai đi. Giờ giải bằng điện thoại thì khó quá

Nhầm. Dùng tam thức bậc 2 chứ

\(\dfrac{a^2}{b+1}+\dfrac{b^2}{c+1}+\dfrac{c^2}{a+1}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+3}=\dfrac{9^2}{9+3}=\dfrac{27}{4}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Chứng minh BĐT \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\) với \(\left(a,b,c>0\right)\)

Trước hết ta cm \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\)\(\Leftrightarrow\frac{x^2b+y^2a}{ab}\ge\frac{x^2+y^2+2xy}{a+b}\)\(\Leftrightarrow\left(x^2b+y^2a\right)\left(a+b\right)\ge ab\left(x^2+y^2+2xy\right)\)(vì tất cả các tử số và mẫu số đều dương)

\(\Leftrightarrow x^2ab+y^2ab+x^2b^2+y^2a^2\ge abx^2+aby^2+2abxy\)\(\Leftrightarrow x^2b^2-2abxy+y^2a^2\ge0\)\(\Leftrightarrow\left(xb-ya\right)^2\ge0\)(luôn đúng)

Vậy BĐT được cm 

Để có đpcm thì ta chỉ cần áp dụng 2 lần BĐT ta vừa chứng minh xong:

\(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y\right)^2}{a+b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)

Nma mik lớp 4

1. Đề thiếu

2. BĐT cần chứng minh tương đương:

\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

Ta có:

\(a^4+b^4+c^4\ge\dfrac{1}{3}\left(a^2+b^2+c^2\right)^2\ge\dfrac{1}{3}\left(ab+bc+ca\right)^2\ge\dfrac{1}{3}.3abc\left(a+b+c\right)\) (đpcm)

3.

Ta có:

\(\left(a^6+b^6+1\right)\left(1+1+1\right)\ge\left(a^3+b^3+1\right)^2\)

\(\Rightarrow VT\ge\dfrac{1}{\sqrt{3}}\left(a^3+b^3+1+b^3+c^3+1+c^3+a^3+1\right)\)

\(VT\ge\sqrt{3}+\dfrac{2}{\sqrt{3}}\left(a^3+b^3+c^3\right)\)

Lại có:

\(a^3+b^3+1\ge3ab\) ; \(b^3+c^3+1\ge3bc\) ; \(c^3+a^3+1\ge3ca\)

\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge3\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^3+b^3+c^3\ge3\)

\(\Rightarrow VT\ge\sqrt{3}+\dfrac{6}{\sqrt{3}}=3\sqrt{3}\)

4.

Ta có:

\(a^3+1+1\ge3a\) ; \(b^3+1+1\ge3b\) ; \(c^3+1+1\ge3c\)

\(\Rightarrow a^3+b^3+c^3+6\ge3\left(a+b+c\right)=9\)

\(\Rightarrow a^3+b^3+c^3\ge3\)

5.

Ta có:

\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\) ; \(\dfrac{a}{b}+\dfrac{c}{a}\ge2\sqrt{\dfrac{c}{b}}\) ; \(\dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\)

\(\Rightarrow\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}+\sqrt{\dfrac{a}{c}}\le\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=1\)

Câu 1:

\(VT=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(VT=1-\dfrac{1}{n}< 1\) (đpcm)