Question

BT1: Cho a,b,c>0. CMR: a²(b+c-a)+b²(c+a-b)+c²(a+b-c)=<3abc

BT2: Cho a,b,c>0. CMR\(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}>=a+b+c\)

BT3: Cho a,b,c>0 thỏa mãn: abc=ab+bc+ca. Chứng minh:

\(\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}=< \dfrac{3}{16}\)

GIÚP MÌNH VỚI. MÌNH ĐANG CẦN GẤP.

Answer 1

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 )

Answer 2

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