Ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=-\dfrac{1}{c^3}\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3+\dfrac{1}{c^3}=0\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=0\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+\dfrac{3}{ab}.\left(-\dfrac{1}{c}\right)=0\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}-\dfrac{3}{abc}=0\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)

Ta có: Điều cần chứng minh là \(A=3abc\) hay \(\dfrac{A}{3abc}=1\)

Thật vậy:

\(\dfrac{A}{3abc}=\left(\dfrac{b^2c^2}{a}+\dfrac{c^2a^2}{b}+\dfrac{a^2b^2}{c}\right).\dfrac{1}{3abc}\)

\(\dfrac{A}{3abc}=\dfrac{b^2c^2}{3a^2bc}+\dfrac{c^2a^2}{3ab^2c}+\dfrac{a^2b^2}{3abc^2}\)

\(\dfrac{A}{3abc}=\dfrac{bc}{3a^2}+\dfrac{ac}{3b^2}+\dfrac{ab}{3c^2}\)

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

\(\dfrac{A}{3abc}=\dfrac{abc}{3}\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{abc}{3}.\dfrac{3}{abc}=1\)

\(\dfrac{A}{3abc}=1\Leftrightarrow A=3abc\left(đpcm\right)\)

a) Áp dụng bất đẳng thức Schur với \(r=1\)

\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)

\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)

\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c\)

b) Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)

Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)

\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c\)

c) Ta có \(abc=ab+bc+ca\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0

\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)

Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)

\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0

\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)

\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)

\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )

Từ ( 1 ) và ( 2 )

\(\Rightarrow VT\le\dfrac{3}{16}\)

\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )

Câu 2 b sos là ra :D

\(BĐT\Leftrightarrow\Sigma_{cyc}\left(\frac{a^3-ab^2}{b^2}\right)\ge0\Leftrightarrow\Sigma_{cyc}\left(\frac{a\left(a-b\right)\left(a+b\right)}{b^2}-2\left(a-b\right)\right)\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\left(a-b\right)\left(\frac{a^2-b^2+ab-b^2}{b^2}\right)\ge0\)\(\Leftrightarrow\Sigma_{cyc}\frac{\left(a-b\right)^2\left(a+2b\right)}{b^2}\ge0\)

BĐT cuối cùng đúng nên ta có Q.E.D.

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

Câu 3 có cách khác ạ,nhưng mà sao em thấy nó sai sai,vì dấu "=" không xảy ra!Mong mọi người check giúp ạ!

gt<=> \(abc=ab+bc+ca\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) (chia hai vế cho abc>0)

Dễ chứng minh \(\frac{1}{36}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{k}\right)\ge\frac{1}{a+b+c+d+e+k}\)

Áp dụng vào,ta có: \(VT\le\frac{6}{36}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{6}< \frac{3}{16}?!?\)

\(A=\frac{b^2c^2}{a}+\frac{c^2a^2}{b}+\frac{a^2b^2}{c}=\frac{a^3b^3+b^3c^3+c^3a^3}{abc}=\frac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{abc}\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)

\(\Rightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3=3.ab.bc.ca=3a^2b^2c^2\)

Vậy \(A=\frac{3a^2b^2c^2}{abc}=3abc\left(a,b,c\ne0\right)\)

\(P=\dfrac{a^2}{ab+\dfrac{1}{b}}+\dfrac{b^2}{bc+\dfrac{1}{c}}+\dfrac{c^2}{ca+\dfrac{1}{a}}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}\)

\(P\ge\dfrac{3\left(ab+bc+ca\right)}{ab+bc+ca+\dfrac{ab+bc+ca}{abc}}=\dfrac{3}{1+\dfrac{1}{abc}}=\dfrac{3abc}{1+abc}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Với a, b, c > 0 có:

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

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

chọn \(\alpha=\dfrac{1}{abc}\Rightarrow dpcm\) 

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

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

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

Theo hệ quả của BĐT AM-GM thì:

\(a^2+b^2+c^2\geq ab+bc+ac\)

\(\Rightarrow \text{VT}\geq \frac{\left ( \frac{ab+bc+ac}{abc} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Ta có đpcm.

Dấu bằng xảy ra khi \(a=b=c\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

=> bc+ac+ab=0

ta có

\(bc+ac=-ab\)

<=> \(\left(bc+ac\right)^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2+2abc^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2-a^2b^2=-2abc^2\)

tương tự

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

\(c^2a^2+a^2b^2-b^2c^2=-2a^2bc\)

thay vào E ta đc

\(E=\dfrac{-a^2b^2c^2}{2ab^2c}-\dfrac{a^2b^2c^2}{2abc^2}-\dfrac{a^2b^2c^2}{2a^2bc}\)

=\(-\dfrac{ac}{2}-\dfrac{ab}{2}-\dfrac{bc}{2}=\dfrac{-\left(ac+ab+bc\right)}{2}=0\) (vì ac+bc+ab=0 cmt)

Áp dụngk BĐt cô-si, ta có 

\(\frac{a^2}{b^2c}+\frac{b^2}{c^2a}+\frac{1}{a}\ge3.\frac{1}{c}\)

Tương tự , rồi cộng vào, ta có 

\(2A+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\Rightarrow A\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left(ĐPCM\right)\)

^_^ 

Bài 1:

Dự đoán dấu "=" xảy ra khi \(a=b=c=1\) ta tính được giá trị là \(9\)

Ta sẽ chứng minh nó là GTLN

Thật vậy ta cần chứng minh

\(\Sigma\dfrac{11a+4b}{4a^2-ab+2b^2}\le\dfrac{3\left(ab+ac+bc\right)}{abc}\)

\(\LeftrightarrowΣ\left(\dfrac{3}{a}-\dfrac{11a+4b}{4a^2-ab+2b^2}\right)\ge0\)

\(\LeftrightarrowΣ\dfrac{\left(a-b\right)\left(a-6b\right)}{a\left(4a^2-ab+2b^2\right)}\ge0\)

\(\LeftrightarrowΣ\left(\dfrac{\left(a-b\right)\left(a-6b\right)}{a\left(4a^2-ab+2b^2\right)}+\dfrac{1}{b}-\dfrac{1}{a}\right)\ge0\)

\(\LeftrightarrowΣ\dfrac{\left(a-b\right)^2\left(a+b\right)}{ab\left(4a^2-ab+2b^2\right)}\ge0\) (luôn đúng)

Bài 2:

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(a^5+b^2+c^2\right)\left(\dfrac{1}{a}+b^2+c^2\right)\ge\left(a^2+b^2+c^2\right)^2\)

\(\Rightarrow\dfrac{1}{a^5+b^2+c^2}\le\dfrac{\dfrac{1}{a}+b^2+c^2}{\left(a^2+b^2+c^2\right)^2}\)

Tương tự rồi cộng theo vế ta có:

\(Σ\dfrac{1}{a^5+b^2+c^2}\le\dfrac{Σ\dfrac{1}{a}+2Σa^2}{\left(a^2+b^2+c^2\right)^2}\)

Ta chứng minh \(Σ\dfrac{1}{a}+2\left(a^2+b^2+c^2\right)\le3\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\) - BĐT cuối đúng

Vậy ta có ĐPCM. Dấu "=" xảy ra khi \(a=b=c=1\)

Bài 3:

Từ \(a+b+c=3abc\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=3\)

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow xy+yz+xz=3\) và BĐT cần chứng minh là

\(x^3+y^3+z^3\ge3\). Áp dụng BĐT AM-GM ta có:

\(x^3+x^3+1\ge3\sqrt[3]{x^3\cdot x^3\cdot1}=3x^2\)

Tương tự có: \(y^3+y^3+1\ge3y^2;z^3+z^3+1\ge3z^2\)

Cộng theo vế 3 BĐT trên ta có:

\(2\left(x^3+y^3+z^3\right)+3\ge3\left(x^2+y^2+z^2\right)\)

Lại có BĐT quen thuộc \(x^2+y^2+z^2\ge xy+yz+xz\)

\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge3\left(xy+yz+xz\right)=9\left(xy+yz+xz=3\right)\)

\(\Rightarrow2\left(x^3+y^3+z^3\right)+3\ge9\Rightarrow2\left(x^3+y^3+z^3\right)\ge6\)

\(\Rightarrow x^3+y^3+z^3\ge3\). BĐT cuối đúng nên ta có ĐPCM

Đẳng thức xảy ra khi \(a=b=c=1\)

T/b:Vâng, rất giỏi

lần sau đăng từng câu 1 dc ko bn :)

system errow

Câu hỏi t/tự