Có A+B = a+b-5-b-c+1 = a-c-4

C-D= b-c-4-b+a = a-c-4

=> A+B=C-D

Ta có: a-c-4 = a-c-4 (hiển nhiên)

         a - c +b - b -5 +1 = a + b -b -4

Mà A=a+b-5   ;   B=-b-c+1

C=b-c-4  ; D=b-a:

         (a + b - 5) +(b - c +1)= (b - c - 4 )+(-b +a )

=> A + B = C + D

Có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=....+2\frac{a+b+c}{abc}=.....\)

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

\(A\ge2\sqrt{\frac{a}{2.2a}}+2\sqrt{\frac{b.2}{2.b}}+\frac{1}{2}.3=\frac{9}{2}\)

Dấu "=" khi \(\left(a;b\right)=\left(1;2\right)\)

\(P=2\left(\frac{4}{2\left(ab+bc+ac\right)}+\frac{1}{a^2+b^2+c^2}\right)+\frac{1}{2\left(ab+bc+ca\right)}\)

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

\(P\ge\frac{18}{\left(a+b+c\right)^2}+\frac{3}{2\left(a+b+c\right)^2}=18+\frac{3}{2}=\frac{39}{2}\)

Dâu "=" khi \(a=b=c=\frac{1}{3}\)

a² + b² \(\ge\frac{1}{2}\)

Lại có \(\frac{a^2+b^2}{2}\) \(\ge\left(\frac{a^2+b^2}{2}\right)^2=\frac{1}{16}\). Suy ra đpcm

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)

Ta chứng minh được

\(a^4+b^4\ge ab\left(a^2+b^2\right)\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

\(\Rightarrow P\le\sum\frac{ab}{ab\left(a^2+b^2\right)+ab}=\sum\frac{1}{a^2+b^2+1}\)

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

Ta lại chứng minh được:

\(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\)

\(\Rightarrow P\le\sum\frac{1}{x^3+y^3+1}\le\sum\frac{xyz}{xy\left(x+y\right)+xyz}=\sum\frac{z}{x+y+z}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Đây là bài thi vào 10 của Thanh Hóa thì phải