\(VT\leΣ\frac{1}{a^2+b^2+1}\le\frac{a^2+b^2+c^2+6}{\left(a+b+c\right)^2}\le\frac{\left(Σa\right)^2}{\left(Σa\right)^2}=1=VP\)

Bạn giải rõ ra được không

\(VT=\Sigma\frac{1}{\frac{a^3}{b}+\frac{b^3}{a}+1}=\Sigma\frac{1}{\frac{a^4}{ab}+\frac{b^4}{ab}+1}\)

Áp dụng BĐT Cauchy-schwar ta có:

\(VT\le\Sigma\frac{1}{\frac{\left(a^2+b^2\right)^2}{2ab}+1}\le\Sigma\frac{1}{\frac{\left(a^2+b^2\right).2ab}{2ab}+1}=\Sigma\frac{1}{a^2+b^2+1}\)\(=\Sigma\frac{c^2+2}{\left(c^2+2\right)\left(a^2+b^2+1\right)}=\Sigma\frac{c^2+2}{\left(a^2c^2+1\right)+\left(b^2c^2+1\right)+\left(a^2+b^2\right)+a^2+b^2+c^2}=\Sigma\frac{c^2+2}{\left(a+b+c\right)^2}=\Sigma\frac{a^2+b^2+c^2+6}{\left(a+b+c\right)^2}\)Áp dụng BĐT AM-GM ta có:

\(ab+bc+ca+ab+bc+ca\ge6.\sqrt[6]{a^4b^4c^4}=6\)

\(\Rightarrow\)\(VT\le\frac{a^2+b^2+c^2+2ab+2bc+2ca}{\left(a+b+c\right)^2}=\frac{\left(\Sigma a\right)^2}{\left(\Sigma a\right)^2}=1\)

Dấu ' = " xảy ra <=> a=b=c

đpcm

\(a^4+b^4\ge\frac{1}{2}\left(a^2+b^2\right)^2=\frac{1}{2}\left(a^2+b^2\right)\left(a^2+b^2\right)\ge ab\left(a^2+b^2\right)\)

\(\Rightarrow VT\le\frac{ab}{ab\left(a^2+b^2\right)+ab}+\frac{bc}{bc\left(b^2+c^2\right)+bc}+\frac{ca}{ca\left(c^2+a^2\right)+ca}\)

\(VT\le\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}\)

Đặt \(\left(a^2;b^2;c^2\right)=\left(x^3;y^3;z^3\right)\Rightarrow xyz=1\)

\(VT\le\frac{1}{x^3+y^3+1}+\frac{1}{y^3+z^3+1}+\frac{1}{z^3+x^3+1}\)

\(VT\le\frac{xyz}{xy\left(x+y\right)+xyz}+\frac{xyz}{yz\left(y+z\right)+xyz}+\frac{xyz}{xz\left(z+x\right)+xyz}\)

\(VT\le\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Sửa đề:  Cho a, b, c là các số thực dương thỏa mãn điều kiện abc=1. Chứng minh rằng

\(\frac{1}{ab+b+2}+\frac{1}{bc+c+2}+\frac{1}{ca+a+2}\le\frac{3}{4}\)

Áp dụng bđt Cauchy-Schwarz ta có:

\(\frac{1}{ab+b+2}=\frac{1}{ab+1+b+1}\le\frac{1}{4}\left(\frac{1}{ab+1}+\frac{1}{b+1}\right)\) \(=\frac{1}{4}\left(\frac{abc}{ab\left(1+c\right)}+\frac{1}{b+1}\right)=\frac{1}{4}\left(\frac{c}{1+c}+\frac{1}{b+1}\right)\)

Tương tự \(\frac{1}{bc+c+2}\le\frac{1}{4}\left(\frac{a}{a+1}+\frac{1}{c+1}\right)\)

          \(\frac{1}{ca+a+2}\le\frac{1}{4}\left(\frac{b}{b+1}+\frac{1}{a+1}\right)\)

Cộng từng vế các bđt trên ta được

\(VT\le\frac{1}{4}\left(\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\right)=\frac{3}{4}\)

Vậy bđt được chứng minh

Dấu "=" xảy ra khi a=b=c=1

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)

lp 8 mà khó thế -,- 

Có \(4=a^4+b^4+c^4+1\ge4\sqrt[4]{\left(abc\right)^4}=4abc\)\(\Leftrightarrow\)\(-abc\ge-1\)

\(\Rightarrow\)\(\frac{1}{4-ab}+\frac{1}{4-bc}+\frac{1}{4-ca}=\frac{a+b+c}{4-abc}\le\frac{a+b+c}{4-1}=\frac{a+b+c}{3}\)

Lại có \(3=a^4+b^4+c^4\ge\frac{\left(a^2+b^2+c^2\right)^2}{3}\ge\frac{\frac{\left(a+b+c\right)^4}{9}}{3}=\frac{\left(a+b+c\right)^4}{27}\)

\(\Leftrightarrow\)\(\left(a+b+c\right)^4\le81\)\(\Leftrightarrow\)\(a+b+c\le3\)

\(\Rightarrow\)\(\frac{1}{4-ab}+\frac{1}{4-bc}+\frac{1}{4-ca}\le\frac{a+b+c}{3}\le\frac{3}{3}=1\) ( đpcm ) 

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)

HSG khổ thế đấy cậu :((

áp dung bđt thức:  1/a +1/b+1/c >= 9/(a+b+c)  (cậu có thể lên mạng tham khảo cách cm bđt này)

           =) điều cần CM (=)  9/(12-ab-bc-ca)

áp dụng bđt thức :  2(ab+bc+ca) =< 2(a²+b²+c²)

triệt tiêu 2 đi rồi ta luôn có một điều hiển nhiên là ab+bc+ca =< a²+b²+c² =< a⁴+b⁴+c⁴ = 3

thay vào dpcm =)  9/(12-ab-bc-ca) =< 9/(12-3) = 1 

dấu "=" xảy ra khi a=b=c=1

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