\(\Leftrightarrow a+b-2\sqrt{ab}>=0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2>=0\)(luôn đúng)

a. Đề bài sai (thực chất là nó đúng 1 cách hiển nhiên nhưng "dạng" thế này nó sai sai vì ko ai cho kiểu này cả)

Ta có: \(abc=ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow abc\ge27\)

\(\Rightarrow a^2+b^2+c^2+5abc\ge a^2+b^2+c^2+5.27>>>>>8\)

b. 

\(4=ab+bc+ca+abc=ab+bc+ca+\sqrt{ab.bc.ca}\le ab+bc+ca+\sqrt{\left(\dfrac{ab+bc+ca}{3}\right)^3}\)

\(\sqrt{\dfrac{ab+bc+ca}{3}}=t\Rightarrow t^3+3t^2-4\ge0\Rightarrow\left(t-1\right)\left(t+2\right)^2\ge0\)

\(\Rightarrow t\ge1\Rightarrow ab+bc+ca\ge3\Rightarrow a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}\ge3\)

- TH1: nếu \(a+b+c\ge4\)

Ta có: \(ab+bc+ca=4-abc\le4\)

\(\Rightarrow P=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)+5abc\ge4^2-2.4+0=8\)

(Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2;2;0\right)\) và các hoán vị)

- TH2: nếu \(3\le a+b+c< 4\)

Đặt \(a+b+c=p\ge3;ab+bc+ca=q;abc=r\)

\(P=p^2-2q+5r=p^2-2q+5\left(4-q\right)=p^2-7q+20\)

Áp dụng BĐT Schur:

\(4=q+r\ge q+\dfrac{p\left(4q-p^2\right)}{9}\Leftrightarrow q\le\dfrac{p^3+36}{4p+9}\)

\(\Rightarrow P\ge p^2-\dfrac{7\left(p^3+36\right)}{4p+9}+20=\dfrac{3\left(4-p\right)\left(p-3\right)\left(p+4\right)}{4p+9}+8\ge8\)

(Dấu "=" xảy ra khi \(a=b=c=1\))

a)\(\dfrac{\left(a+b\right)^2}{4}\ge ab\)\(\Leftrightarrow\dfrac{a^2+ab+b^2}{4}\ge0\)\(\Leftrightarrow\dfrac{\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}}{4}\ge0\left(đpcm\right)\)

Vậy \(\dfrac{a+b}{2}\ge\sqrt{ab}\)

b) Áp dụng Cauchy, ta có:

\(\dfrac{bc}{a}+\dfrac{ca}{b}\ge2\sqrt{\dfrac{bc}{a}.\dfrac{ca}{b}}=2c\)

Tương tự: \(\dfrac{ca}{b}+\dfrac{ab}{c}\ge2a\)

\(\dfrac{ab}{c}+\dfrac{bc}{a}\ge2b\)

Cộng vế theo vế các BĐT vừa chứng minh rồi rút gọn ta được đpcm.

c) ta có \(3a+5b=12\Rightarrow a=\dfrac{12-5b}{3}=4-\dfrac{5b}{3}\)

\(\Rightarrow P=ab=\left(4-\dfrac{5b}{3}\right)b=4b-\dfrac{5b^2}{3}\)

\(\Rightarrow15P=60b-25b^2=36-\left(25b^2-60b+36\right)=36-\left(5b-6\right)^2\)

\(\Rightarrow15P\le36\Rightarrow P\le\dfrac{36}{15}=\dfrac{12}{5}\) Vậy GTLN của \(P=\dfrac{12}{5}\) tại \(a=2;b=\dfrac{6}{5}\)

Ta có

\(a^4+b^4+c^4-abc\left(a+b+c\right)=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)-abc\left(a+b+c\right)\)

\(=\left(a^2+b^2+c^2\right)^2-2\left[\left(ab+bc+ac\right)^2-2a^2bc-2ab^2c-2abc^2\right]-a^2bc-ab^2c-abc^2\)

\(=\left(a^2+b^2+c^2\right)^2-2\left(ab+bc+ac\right)^2+4a^2bc+4ab^2c+4abc^2-a^2bc-ab^2c-abc^2\)

\(=\left[\left(a+b+c\right)^2-2\left(ab+bc+ac\right)\right]^2-2\left(ab+bc+ac\right)^2+abc\left(4a+4b+4c-a-b-c\right)\)

\(=\left(a+b+c\right)^4-2\left(a+b+c\right)^2.2\left(ab+bc+ac\right)+4\left(ab+bc+ca\right)^2-2\left(ab+bc+ac\right)^2+abc\left(3a+3b+3c\right)\)

\(=\left(a+b+c\right)^4-4\left(a+b+c\right)^2\left(ab+bc+ca\right)+2\left(ab+bc+ac\right)^2+3abc\ge0\)

Ap dung BDt co si ta co

\(a^4+b^4\ge2a^2b^2\)

\(b^4+c^4\ge2b^2c^2\)

\(c^4+a^4\ge2a^2c^2\)

=> \(a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\)(1)

Lai co \(a^2b^2+b^2c^2\ge2ab^2c\)

          \(b^2c^2+c^2a^2\ge2abc^2\)

          \(c^2a^2+a^2b^2\ge2a^2bc\)

=> \(a^2b^2+b^2c^2+c^2a^2\ge abc\left(a+b+c\right)\)(2)

Từ (1) va (2) => \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

Bạn Nguyễn Huệ Lam sai rồi chưa chắc

\(\left(a+b+c\right)^4-4\left(a+b+c\right)^2\left(ab+bc+ca\right)\)\(\ge0\)

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

\(\Leftrightarrow\left(1+ab\right)\left(2+a^2+b^2\right)\ge2a^2b^2+2a^2+2b^2+2\)

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

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)

b/ \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{2}{1+b^4}\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^4}\ge\frac{4}{1+ab^3}\)

\(\Rightarrow\frac{1}{1+a^4}+\frac{3}{1+b^4}\ge\frac{4}{1+ab^3}\)

Hoàn toàn tương tự: \(\frac{1}{1+b^4}+\frac{3}{1+c^4}\ge\frac{4}{1+bc^3}\); \(\frac{1}{1+c^4}+\frac{3}{1+a^4}\ge\frac{4}{1+a^3c}\)

Cộng vế với vế ta có đpcm

a) Giả sử:

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

\(\Rightarrow\frac{a^2+2ab+b^2}{4}\ge ab\)

\(\Rightarrow\frac{a^2+2ab+b^2}{4}-ab\ge0\)

\(\Rightarrow\frac{\left(a-b\right)^2}{4}\ge0\Rightarrow\left(a-b\right)^2\ge0\) (luôn đúng )

=> đpcm

b, Bất đẳng thức Cauchy cho các cặp số dương \(\frac{bc}{a}\)và \(\frac{ca}{b};\frac{bc}{a}\)và \(\frac{ab}{c};\frac{ca}{b}\)và \(\frac{ab}{c}\)

Ta lần lượt có : \(\frac{bc}{a}+\frac{ca}{b}\ge\sqrt[2]{\frac{bc}{a}.\frac{ca}{b}}=2c;\frac{bc}{a}+\frac{ab}{c}\ge\sqrt[2]{\frac{bc}{a}.\frac{ab}{c}}=2b;\frac{ca}{b}+\frac{ab}{c}\ge\sqrt[2]{\frac{ca}{b}.\frac{ab}{c}}\)

Cộng từng vế ta đc bất đẳng thức cần chứng minh . Dấu ''='' xảy ra khi \(a=b=c\)

c, Với các số dương \(3a\) và \(5b\), Theo bất đẳng thức Cauchy ta có \(\frac{3a+5b}{2}\ge\sqrt{3a.5b}\)

\(\Leftrightarrow\left(3a+5b\right)^2\ge4.15P\)( Vì \(P=a.b\)) 

\(\Leftrightarrow12^2\ge60P\)\(\Leftrightarrow P\le\frac{12}{5}\Rightarrow maxP=\frac{12}{5}\)

Dấu ''='' xảy ra khi \(3a=5b=12:2\)

\(\Leftrightarrow a=2;b=\frac{6}{5}\)

a.

Xét hiệu:

\(a^3+b^3-ab\left(a+b\right)=\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\)

\(=a^2-ab+b^2-ab=a^2-2ab+b^2\)

\(=\left(a-b\right)^2\ge0\)

=> BĐT luôn đúng

b.

Xét hiệu:

\(a^4+b^4-a^3b-ab^3=\left(a^4-a^3b\right)-\left(b^4-ab^3\right)\)

\(=a^3\left(a-b\right)-b^3\left(a-b\right)=\left(a^3-b^3\right)\left(a-b\right)\)

\(=\left(a-b\right)\left(a^2+ab+b^2\right)\left(a-b\right)\)

\(=\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

=> BĐT luôn đúng

a)

\(a^3+b^3\ge ab\left(a+b\right)\forall a,b>0\)

\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)

\(\Rightarrow a^2-ab+b^2\ge ab\)

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

\(\Rightarrowđpcm\)

b)

\(a^4+b^4\ge a^3b+ab^3\)

\(\Leftrightarrow a^4-ab^3+b^4-a^3b\ge0\)

\(\Leftrightarrow a\left(a^3-b^3\right)-b\left(a^3-b^3\right)\ge0\)

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

\(\Rightarrowđpcm\)

c)

\(\left(a+1\right)\left(b+1\right)\ge\left(\sqrt{ab}+1\right)^2\)

\(\Leftrightarrow\left(a+1\right)\left(b+1\right)-\left(\sqrt{ab}+1\right)^2\ge0\)

\(\Leftrightarrow1+b+a+ab-ab-2\sqrt{ab}-1\ge0\)

\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)

Dấu bằng xảy ra khi \(a=b\)

d)

\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ac\)

Áp dụng bất đẳng thức AM-GM ta được

\(\dfrac{a^3}{b}+ab\ge2\sqrt{\dfrac{a^3}{b}.ab}\)

\(\Leftrightarrow\dfrac{a^3}{b}+ab\ge2a^2\)

Tương tự ta được

\(\dfrac{b^3}{c}+bc\ge2b^2,\dfrac{c^3}{a}+ac\ge2c^2\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ac\ge2\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ac\right)\)

Mặt khác ta có:\(a^2+b^2+c^2\ge ab+bc+ac\) (hệ quả bất đẳng thức AM-GM)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ac\left(đpcm\right)\)

Dấu bằng xảy ra khi \(x=y=z;x,y,z>0\)

\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(=1+\dfrac{a}{b}+1+\dfrac{b}{a}=2+\dfrac{a}{b}+\dfrac{b}{a}\)

Áp dụng BĐT Cauchy cho 2 số dương \(\dfrac{a}{b}\) và \(\dfrac{b}{a}\), ta có:

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

Vậy: \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)

Dấu ''='' xảy ra khi và chỉ khi a=b

P/S: Nếu chưa học Cauchy thì xét hằng đẳng thức \(\left(a-b\right)^2\ge0\Rightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

Toán lớp 8

Ta bien doi BDT can chung minh

\(a+b\ge\frac{4ab}{1+ab}\)

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

\(\Leftrightarrow a+\frac{1}{a}+b+\frac{1}{b}\ge4\)

Ta co:

\(a+\frac{1}{a}\ge2\)

\(b+\frac{1}{b}\ge2\)

\(\Rightarrow a+\frac{1}{a}+b+\frac{1}{b}\ge4\)

Dau '=' xay ra khi \(a=b=1\)