3/ Áp dụng bất đẳng thức AM-GM, ta có :

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)

\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)

\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)

Cộng 3 vế của BĐT trên ta có :

\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)

\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)

Bài 1:

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)

Do đó:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Bài 2:

Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)

\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)

\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)

Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

\(\Leftrightarrow\left(1+abc\right)\left(\dfrac{1}{a\left(1+b\right)}+\dfrac{1}{b\left(1+c\right)}+\dfrac{1}{c\left(1+a\right)}\right)\ge3\)

Ta có:

\(\left(1+abc\right).\dfrac{1}{a\left(1+b\right)}=\dfrac{1+abc}{a+ab}=\dfrac{1+a+ab+abc-a-ab}{a+ab}=\dfrac{1+a}{a\left(1+b\right)}+\dfrac{b\left(1+c\right)}{1+b}-1\)

\(\Rightarrow VT=\dfrac{1+a}{a\left(1+b\right)}+\dfrac{b\left(1+c\right)}{1+b}+\dfrac{1+b}{b\left(1+c\right)}+\dfrac{c\left(1+a\right)}{1+c}+\dfrac{1+c}{c\left(1+a\right)}+\dfrac{a\left(1+b\right)}{1+a}-3\)

\(VT\ge6\sqrt[6]{\dfrac{abc\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}{abc\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}}-3=3\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

1.

BĐT cần chứng minh tương đương:

\(\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\)

Ta có:

\(\left(ab-1\right)^2=a^2b^2-2ab+1=a^2b^2-a^2-b^2+1+a^2+b^2-2ab\)

\(=\left(a^2-1\right)\left(b^2-1\right)+\left(a-b\right)^2\ge\left(a^2-1\right)\left(b^2-1\right)\)

Tương tự: \(\left(bc-1\right)^2\ge\left(b^2-1\right)\left(c^2-1\right)\)

\(\left(ca-1\right)^2\ge\left(c^2-1\right)\left(a^2-1\right)\)

Do \(a;b;c\ge1\)  nên 2 vế của các BĐT trên đều không âm, nhân vế với vế:

\(\left[\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\right]^2\ge\left[\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\right]^2\)

\(\Rightarrow\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Câu 2 em kiểm tra lại đề có chính xác chưa

2.

Câu 2 đề thế này cũng làm được nhưng khá xấu, mình nghĩ là không thể chứng minh bằng Cauchy-Schwaz được, phải chứng minh bằng SOS

Không mất tính tổng quát, giả sử \(c=max\left\{a;b;c\right\}\)

\(\Rightarrow\left(c-a\right)\left(c-b\right)\ge0\) (1)

BĐT cần chứng minh tương đương:

\(\dfrac{1}{a}-\dfrac{a+b}{bc+a^2}+\dfrac{1}{b}-\dfrac{b+c}{ac+b^2}+\dfrac{1}{c}-\dfrac{c+a}{ab+c^2}\ge0\)

\(\Leftrightarrow\dfrac{b\left(c-a\right)}{a^3+abc}+\dfrac{c\left(a-b\right)}{b^3+abc}+\dfrac{a\left(b-c\right)}{c^3+abc}\ge0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)+a\left(c-b\right)}{a^3+abc}+\dfrac{c\left(a-b\right)}{b^3+abc}+\dfrac{a\left(b-c\right)}{c^3+abc}\ge0\)

\(\Leftrightarrow c\left(b-a\right)\left(\dfrac{1}{a^3+abc}-\dfrac{1}{b^3+abc}\right)+a\left(c-b\right)\left(\dfrac{1}{a^3+abc}-\dfrac{1}{c^3+abc}\right)\ge0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)\left(b^3-a^3\right)}{\left(a^3+abc\right)\left(b^3+abc\right)}+\dfrac{a\left(c-b\right)\left(c^3-a^3\right)}{\left(a^3+abc\right)\left(c^3+abc\right)}\ge0\)

\(\Leftrightarrow\dfrac{c\left(b-a\right)^2\left(a^2+ab+b^2\right)}{\left(a^3+abc\right)\left(b^3+abc\right)}+\dfrac{a\left(c-b\right)\left(c-a\right)\left(a^2+ac+c^2\right)}{\left(a^3+abc\right)\left(c^3+abc\right)}\ge0\)

Đúng theo (1)

Dấu "=" xảy ra khi \(a=b=c\)

Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow xyz=1\)

\(P=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)

Áp dụng BĐT AM - GM ta có:

$ \frac{a^3}{(1 + b)(1 + c)} + \frac{1 + b}{8} + \frac{1 + c}{8} \geq \frac{3}{4}a$

$\frac{b^3}{(1 + c)(1 + a)} + \frac{1 + c}{8} + \frac{1 + a}{8} \geq \frac{3}{4}b$

$\frac{c^3}{(1 + a)(1 + b)} + \frac{1 + a}{8} + \frac{1 + b}{8} \geq \frac{3}{4}c $

Cộng vế theo vế ta được:

$ P + \frac{2(a + b + c) + 6}{8} \geq \frac{3}{4}(a + b + c) $

$<=> P \geq \frac{1}{2}(a + b + c) - \frac{3}{4}$

$=> P \geq \frac{3}{4} (dpcm)$

Lời giải:

Áp dụng hệ quả của BĐT AM-GM:

\(\text{VT}^2=\left[\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\right]^2\geq 3\left(\frac{1}{ab(a+1)(b+1)}+\frac{1}{bc(b+1)(c+1)}+\frac{1}{ca(a+1)(c+1)}\right)\)

\(\Leftrightarrow \text{VT}^2\geq 3.\frac{a^2+b^2+c^2+a+b+c}{abc(a+1)(b+1)(c+1)}\geq 3.\frac{a+b+c+ab+bc+ac}{abc(a+1)(b+1)(c+1)}\)

\(\Leftrightarrow \text{VT}^2\geq \frac{3}{abc}-\frac{3(abc+1)}{abc(a+1)(b+1)(c+1)}\) \((1)\)

Ta sẽ cm \((a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3\). Thật vậy:

Áp dụng BĐT AM-GM:

\(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}\)

\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}\)

Cộng theo vế: \(\Rightarrow 3\geq \frac{3(\sqrt[3]{abc}+1)}{\sqrt[3]{(a+1)(b+1)(c+1)}}\)

\(\Rightarrow (a+1)(b+1)(c+1)\geq (\sqrt[3]{abc}+1)^3\) (2)

Từ \((1),(2)\Rightarrow \text{VT}^2\geq \frac{3}{abc}-\frac{3(abc+1)}{abc(1+\sqrt[3]{abc})^3}=\frac{9}{\sqrt[3]{a^2b^2c^2}(1+\sqrt[3]{abc})^2}=\text{VP}^2\)

\(\Leftrightarrow \text{VT}\geq \text{VP}\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c=1\)

ap dung bdt holder

Bài 1:

Áp dụng BĐT AM-GM ta có:

$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}$

$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}$

Cộng theo vế và thu gọn:

$\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$

$\Leftrightarrow 3\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$

$\Rightarrow (a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3$

Ta có đpcm.

Bài 2:

$a^3+a^3+a^3+a^3+b^3+c^3\geq 6\sqrt[6]{a^{12}b^3c^3}=6a^2\sqrt{bc}$

$b^3+b^3+b^3+b^3+a^3+c^3\geq 6b^2\sqrt{ac}$

$c^3+c^3+c^3+c^3+a^3+b^3\geq 6c^2\sqrt{ab}$

Cộng theo vế và rút gọn thu được:

$a^3+b^3+c^3\geq a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}$ 

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=c$

Bài 3:

Áp dụng BĐT Cauchy-Schwarz:

$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{(a+b+c)^2}{b+c+c+a+a+b}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$

\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(a+c\right)}+\dfrac{1}{c^3\left(a+b\right)}\)

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

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

\(=\dfrac{b^2c^2}{a^2bc\left(b+c\right)}+\dfrac{a^2c^2}{ab^2c\left(a+c\right)}+\dfrac{a^2b^2}{abc^2\left(a+b\right)}\)

\(Cauchy-Schwarz:\)

\(VT\ge\dfrac{\left(bc+ac+ab\right)^2}{abc\left[a\left(b+c\right)+b\left(a+c\right)+c\left(a+b\right)\right]}\)

\(=\dfrac{\left(bc+ac+ab\right)^2}{2\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2}\)

\(AM-GM:\)

\(ab+bc+ca\ge\sqrt[3]{\left(abc\right)^2}=3\)

\(\Rightarrow VT\ge\dfrac{ab+bc+ca}{2}\ge\dfrac{3}{2}\)

\("="\Leftrightarrow a=b=c=1\)

Lời giải khác:

Áp dụng BĐT AM-GM:

\(\frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}\geq 2\sqrt{\frac{1}{4a^2}}=\frac{1}{a}=\frac{abc}{a}=bc\)

\(\frac{1}{b^3(a+c)}+\frac{b(a+c)}{4}\geq 2\sqrt{\frac{1}{4b^2}}=\frac{1}{b}=\frac{abc}{b}=ac\)

\(\frac{1}{c^3(a+b)}+\frac{c(a+b)}{4}\geq 2\sqrt{\frac{1}{4c^2}}=\frac{1}{c}=\frac{abc}{c}=ab\)

Cộng theo vế và rút gọn:

\(\Rightarrow \frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}+\frac{ab+bc+ac}{2}\ge ab+bc+ac\)

\(\Rightarrow \frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\geq \frac{ab+bc+ac}{2}\geq \frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\) (AM_GM)

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c=1$

wow

wow, chắc xu học lớp 9

\(\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{a^3\left(a+b\right)\left(b+c\right)}{64}}=\dfrac{3a}{4}\)

Tương tự:

\(\dfrac{b^3}{\left(b+c\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{c+a}{8}\ge\dfrac{3b}{4}\)

\(\dfrac{c^3}{\left(c+a\right)\left(a+b\right)}+\dfrac{c+a}{8}+\dfrac{a+b}{8}\ge\dfrac{3c}{4}\)

Cộng vế:

\(VT+\dfrac{4\left(a+b+c\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)

\(\Rightarrow VT\ge\dfrac{a+b+c}{4}\)

Dấu "=" xảy ra khi \(a=b=c\)