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)

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 !!!

Ta viết lại bất đẳng thức cần chứng minh thành\(\sqrt{\frac{2\left(a+3\right)}{a+bc}}+\sqrt{\frac{2\left(b+3\right)}{b+ca}}+\sqrt{\frac{2\left(c+3\right)}{c+ab}}\ge6\)

Theo giả thiết, ta có a + b + c = 3 nên\(\sqrt{\frac{2\left(a+3\right)}{a+bc}}=\sqrt{\frac{2\left(a+a+b+c\right)}{a+bc}}=\sqrt{2\left(\frac{a+b}{a+bc}+\frac{a+c}{a+bc}\right)}\)\(\ge\sqrt{\frac{a+b}{a+bc}}+\sqrt{\frac{a+c}{a+bc}}\)(Áp dụng bất đẳng thức \(\sqrt{2\left(x+y\right)}\ge\sqrt{x}+\sqrt{y}\))

Hoàn toàn tương tự, ta được: \(\sqrt{\frac{2\left(b+3\right)}{b+ca}}\ge\sqrt{\frac{b+a}{b+ca}}+\sqrt{\frac{b+c}{b+ca}}\); \(\sqrt{\frac{2\left(c+3\right)}{c+ab}}\ge\sqrt{\frac{c+a}{c+ab}}+\sqrt{\frac{c+b}{c+ab}}\)

Cộng theo vế ba bất đẳng thức trên, ta được: \(\sqrt{\frac{2\left(a+3\right)}{a+bc}}+\sqrt{\frac{2\left(b+3\right)}{b+ca}}+\sqrt{\frac{2\left(c+3\right)}{c+ab}}\)\(\ge\sqrt{\frac{a+b}{a+bc}}+\sqrt{\frac{a+c}{a+bc}}+\sqrt{\frac{b+a}{b+ca}}+\sqrt{\frac{b+c}{b+ca}}+\sqrt{\frac{c+a}{c+ab}}+\sqrt{\frac{c+b}{c+ab}}\)

Áp dụng bất đẳng thức Bunyakovsky dạng phân thức, ta được: \(\sqrt{\frac{a+b}{a+bc}}+\sqrt{\frac{a+b}{b+ca}}\ge\frac{4\sqrt{a+b}}{\sqrt{a+bc}+\sqrt{b+ca}}\ge\frac{2\sqrt{2}\sqrt{a+b}}{\sqrt{a+bc+b+ca}}=\frac{2\sqrt{2}}{\sqrt{c+1}}\)(*)

Tương tự ta có: \(\sqrt{\frac{b+c}{b+ca}}+\sqrt{\frac{b+c}{c+ab}}\ge\frac{2\sqrt{2}}{\sqrt{a+1}}\)(**) ; \(\sqrt{\frac{c+a}{c+ab}}+\sqrt{\frac{c+a}{a+bc}}\ge\frac{2\sqrt{2}}{\sqrt{b+1}}\)(***)

Cộng theo vế ba bất đẳng thức (*), (**) và (***) suy ra \(\sqrt{\frac{a+b}{a+bc}}+\sqrt{\frac{a+c}{a+bc}}+\sqrt{\frac{b+a}{b+ca}}+\sqrt{\frac{b+c}{b+ca}}+\sqrt{\frac{c+a}{c+ab}}+\sqrt{\frac{c+b}{c+ab}}\)\(\ge\frac{2\sqrt{2}}{\sqrt{c+1}}+\frac{2\sqrt{2}}{\sqrt{a+1}}+\frac{2\sqrt{2}}{\sqrt{b+1}}\)

Do đó ta có: \(\sqrt{\frac{2\left(a+3\right)}{a+bc}}+\sqrt{\frac{2\left(b+3\right)}{b+ca}}+\sqrt{\frac{2\left(c+3\right)}{c+ab}}\ge\frac{2\sqrt{2}}{\sqrt{c+1}}+\frac{2\sqrt{2}}{\sqrt{a+1}}+\frac{2\sqrt{2}}{\sqrt{b+1}}\)

Phép chứng minh sẽ hoàn tất nếu ta chỉ ra được \(\frac{2\sqrt{2}}{\sqrt{c+1}}+\frac{2\sqrt{2}}{\sqrt{a+1}}+\frac{2\sqrt{2}}{\sqrt{b+1}}\ge6\)hay \(\frac{1}{\sqrt{c+1}}+\frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+1}}\ge\frac{3}{\sqrt{2}}\)

Thật vậy, áp dụng bất đẳng thức Cauchy – Schwarz ta được \(\frac{1}{\sqrt{c+1}}+\frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+1}}\ge\frac{9}{\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}}\ge\frac{9}{\sqrt{3\left(a+b+c+3\right)}}=\frac{3}{\sqrt{2}}\)

Vậy bất đẳng thức được chứng minh 

Đẳng thức xảy ra khi a = b = c = 1

132-79=

ta có :

\(\frac{a^3+b^3}{a^2+ab+b^2}=\frac{2a^3}{a^2+ab+b^2}+\frac{b^3-a^3}{a^2+ab+b^2}=\frac{2a^3}{a^2+ab+b^3}+b-a\)

tương tự rồi cộng theo vế : 

\(LHS\ge2\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)

áp dụng bđt cô si

 \(\frac{a^3}{a^2+ab+b^2}+\frac{a^2+ab+b^2}{9}+\frac{1}{3}\ge\frac{3a}{3}=a\)

tương tự rồi cộng theo vế 

\(2\left(\frac{a^3}{a^2+ab+b^2}+...\right)\ge a+b+c-1-\frac{2\left(a^2+b^2+c^2+ab+bc+ca\right)}{9}\)

\(\ge\frac{2\left(9-a^2-b^2-c^2-ab-bc-ca\right)}{9}\)

đến đây chịu :)))))

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

Ta có BĐT phụ: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)( cái này nhân chéo lên tự cm nha )

\(\Rightarrow\frac{a^3+b^3}{a^2+ab+b^2}\ge\frac{1}{3}\left(a+b\right)\)

CMTT: \(\frac{b^3+c^3}{b^2+bc+c^2}\ge\frac{1}{3}\left(b+c\right);\frac{c^3+a^3}{c^2+ca+a^2}\ge\frac{1}{3}\left(c+a\right)\)

\(\Rightarrow VT\ge\frac{2}{3}\left(a+b+c\right)\ge\frac{2}{3}.3\sqrt[3]{abc}=2\left(đpcm\right)\)

\(A=\frac{\frac{1}{2}a^2\left(\sqrt[3]{b}+\sqrt[3]{c}+1\right)\left[\left(\sqrt[3]{b}-\sqrt[3]{c}\right)^2+\left(\sqrt[3]{b}-1\right)^2+\left(\sqrt[3]{c}-1\right)^2\right]}{2\left(a+2\right)\left(a+\sqrt[3]{bc}\right)}\ge0\)

\(\Sigma_{cyc}\frac{a^2}{a+\sqrt[3]{bc}}=\Sigma_{cyc}A+\Sigma_{cyc}\frac{2\left(a-1\right)^2}{3\left(a+2\right)}+\frac{5}{6}\left(a+b+c\right)-1\ge\frac{5}{6}\left(a+b+c\right)-1=\frac{3}{2}\)

Áp dụng bất đẳng thức cộng mẫu số 

\(\Rightarrow\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\)\(\ge\frac{\left(a+b+c\right)^2}{a+b+c+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)

\(\Rightarrow\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\)\(\ge\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)

Chứng minh rằng : \(\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\frac{3}{2}\)

\(\Leftrightarrow18\ge3\left(3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}\right)\)

\(\Leftrightarrow18\ge9+3\sqrt[3]{bc}+3\sqrt[3]{ca}+3\sqrt[3]{ab}\)

\(\Leftrightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)

Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm

\(\Rightarrow\hept{\begin{cases}a+b+1\ge3\sqrt[3]{ab}\\b+c+1\ge3\sqrt[3]{bc}\\c+a+1\ge3\sqrt[3]{ca}\end{cases}}\)

\(\Rightarrow2\left(a+b+c\right)+3\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)

\(\Rightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\left(đpcm\right)\)

Vì \(\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\frac{3}{2}\)

Mà \(\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\ge\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)

\(\Rightarrow\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\ge\frac{3}{2}\left(đpcm\right)\)

Chúc bạn học tốt !!!

Vào thống kê hỏi đáp xem nhé. Bài này chỉ cần biểu diễn dưới dạng tổng bình phương là xong.

ta có \(\frac{a^3}{b^2+3}+\frac{b^3}{c^2+3}+\frac{c^3}{a^2+3}\ge\frac{3}{4}\) (***)

do ab+bc+ca=3 nên

VT (***)=\(\frac{a^3}{b^2+ab+bc+ca}+\frac{b^3}{c^2+ab+bc+ca}+\frac{c^3}{a^2+ab+bc+ca}\)

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

áp dụng bđt AM-GM ta có \(\frac{a^3}{\left(b+c\right)\left(c+a\right)}+\frac{b+c}{8}+\frac{a+b}{8}\ge\frac{3a}{4}\)

\(\Rightarrow\frac{a^3}{\left(b+c\right)\left(c+a\right)}\ge\frac{5a-2b-c}{8}\left(1\right)\)

chứng minh tương tự ta cũng được

\(\hept{\begin{cases}\frac{b^3}{\left(c+a\right)\left(a+b\right)}\ge\frac{5b-2c-a}{8}\left(2\right)\\\frac{c^3}{\left(a+b\right)\left(c+a\right)}\ge\frac{5c-2a-b}{8}\left(3\right)\end{cases}}\)

cộng theo vế với vế của (1),(2) và (3) ta được VT (***) \(\ge\frac{a+b+c}{4}\)

mặt khác ta dễ dàng chứng minh được \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}=3\)

đẳng thức xảy ra khi a=b=c=1 (đpcm)

ok cảm ơn bạn nha :)