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Sửa \(\dfrac{1}{3}\rightarrow3\)
Từ \(a+b+c+ab+bc+ca=6abc\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=6\)
Ta có: \(\dfrac{1}{a^2}+1\ge\dfrac{2}{a};\dfrac{1}{b^2}+1\ge\dfrac{2}{b};\dfrac{1}{c^2}+1\ge\dfrac{2}{c}\)
Và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge\dfrac{2}{ab};\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{2}{bc};\dfrac{1}{c^2}+\dfrac{1}{a^2}\ge\dfrac{2}{ac}\)
Cộng theo vế các BĐT trên ta có:
\(3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\Leftrightarrow3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\right)\ge12\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\ge4\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge3\)
\("="\Leftrightarrow a=b=c=1\)
Bài 1:
Áp dụng BĐT Holder:
\((a^7+b^7+c^7)(a+b+c)(a+b+c)\geq (a^3+b^3+c^3)^3\)
\(\Rightarrow P=a^7+b^7+c^7\geq \frac{(a^3+b^3+c^3)^3}{(a+b+c)^2}\) \((1)\)
Tiếp tục Holder:
\((a^3+b^3+c^3)(1+1+1)(1+1+1)\geq (a+b+c)^3\)
\(\Rightarrow (a+b+c)\leq \sqrt[3]{9(a^3+b^3+c^3)}\) \((2)\)
Từ \((1),(2)\Rightarrow P\geq \frac{\sqrt[3]{(a^3+b^3+c^3)^7}}{\sqrt[3]{81}}\) \((3)\)
Áp dụng BĐT AM-GM:
\((a^3+b^3+c^3)^2\geq 3(a^3b^3+b^3c^3+c^3a^3)\geq 3\)
\(\Rightarrow a^3+b^3+c^3\geq \sqrt{3}\) \((4)\)
Từ \((3),(4)\Rightarrow P\geq \sqrt[6]{\frac{1}{3}}\)
Vậy \(P_{\min}=\sqrt[6]{\frac{1}{3}}\Leftrightarrow a=b=c=\sqrt[6]{\frac{1}{3}}\)
Bài 2:
Áp dụng BĐT AM-GM:
\(a^3+\sqrt{\frac{1}{27}}+\sqrt{\frac{1}{27}}\geq 3\sqrt[3]{a^3.\sqrt{\frac{1}{27^2}}}=a\)
\(b^3+\sqrt{\frac{1}{27}}+\sqrt{\frac{1}{27}}\geq 3\sqrt[3]{b^3.\sqrt{\frac{1}{27^2}}}=b\)
\(c^3+\sqrt{\frac{1}{27}}+\sqrt{\frac{1}{27}}\geq 3\sqrt[3]{c^3.\sqrt{\frac{1}{27^2}}}=c\)
Cộng theo vế:
\(a^3+b^3+c^3+6\sqrt{\frac{1}{27}}\geq a+b+c\)
Áp dụng BĐT AM-GM:
\((a+b+c)^2\geq 3(ab+bc+ac)=3\Rightarrow a+b+c\geq \sqrt{3}\)
Do đó, \(a^3+b^3+c^3\geq \sqrt{3}-6\sqrt{\frac{1}{27}}=\sqrt{\frac{1}{3}}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=\sqrt{\frac{1}{3}}\)
a.
\(a^2+b^2+c^2\ge ab+bc+ca\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
(luôn đúng)
b. Áp dụng BĐT \(x^2+y^2\ge2xy\)
\(a^2+b^2\ge2ab,a^2+1\ge2a,b^2+1\ge2b\)\(\Rightarrow2\left(a^2+b^2+1\right)\ge2\left(ab+a+b\right)\Leftrightarrow a^2+b^2+1\ge ab+a+b\)
c. Tương tự câu b
Áp dụng BĐT Cô si ta có
i. \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}},\frac{1}{b}+\frac{1}{c}\ge\frac{2}{\sqrt{bc}},\frac{1}{c}+\frac{1}{a}\ge\frac{2}{\sqrt{ca}}\)
\(\Rightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge2\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)\)\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\)
k. Tương tự câu i
a) Áp dụng BĐT Cauchy-Schwarz dạng Engel: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Tương tự:\(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c};\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)
Cộng theo vế 3 BĐT trên rồi chia cho 2 ta thu được đpcm
Đẳng thức xảy ra khi \(a=b=c\)
b)Đặt \(a+b=x;b+c=y;c+a=z\). Cần chứng minh:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
Cách làm tương tự câu a.
c) \(VT=\Sigma_{cyc}\frac{1}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\Sigma_{cyc}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\le\frac{1}{16}\Sigma\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
Đẳng thức xảy ra khi \(a=b=c=\frac{3}{4}\)
d) Em làm biếng quá anh làm nốt đi:P
Câu d : \(p=\frac{a+b+c}{2}\Rightarrow2p=a+b+c\)
Ta có : \(\frac{1}{p-a}+\frac{1}{p-b}\ge\frac{4}{2p-a-b}=\frac{4}{c}\)
Tương tự : \(\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{4}{a}\)
\(\frac{1}{p-c}+\frac{1}{p-a}\ge\frac{4}{b}\)
\(\Leftrightarrow2\left(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\right)\ge\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)
\(\Leftrightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\left(đpcm\right)\)
Dấu \("="\) xảy ra khi nó là tam giác đều
\(\frac{a^2+b^2}{a-b}=\frac{a^2+b^2-2ab+2ab}{a-b}=\frac{\left(a-b\right)^2}{a-b}+\frac{2}{a-b}=a-b+\frac{2}{a-b}\ge2\sqrt{\frac{2\left(a-b\right)}{a-b}}=2\sqrt{2}\)
Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}ab=1\\a-b=\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{\sqrt{6}+\sqrt{2}}{2}\\b=\frac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)
\(\sqrt{\frac{ab+2c^2}{1+ab-c^2}}=\sqrt{\frac{ab+2c^2}{a^2+b^2+ab}}=\frac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(a^2+b^2+ab\right)}}\ge\frac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\ge\frac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự: \(\sqrt{\frac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\) ; \(\sqrt{\frac{ca+2b^2}{1+ac-b^2}}\ge ca+2b^2\)
Cộng vế với vế:
\(VT\ge2\left(a^2+b^2+c^2\right)+ab+bc+ca=2+ab+bc+ca\)
Bài này rất dài dòng nhưng cũng rất quen.
https://diendantoanhoc.net/topic/153766-bổ-đề-hoán-vị/
Bích Ngọc Huỳnh & erone - anotherway
Ta sẽ tìm hàm số \(f\left(q\right)\) sao cho
\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge f(q) \forall a,b,c>0.\)
\(\Leftrightarrow \sum \dfrac{a}{b}+\sum \dfrac{b}{a}\ge 2f(q)+\sum \dfrac{b}{a}-\sum \dfrac{a}{b} \)
Or \(\sum ab(a+b)-2abc\cdot f(q)\ge (a-b)(b-c)(c-a)\)
Need to pro \(\sum ab(a+b)-2abc\cdot f(q)\ge \sqrt{(a-b)^2(b-c)^2(c-a)^2}.\)
Đặt \(p=a+b+c,q=ab+bc+ca,r=abc\)
\((pq-3r)-2f(q)\cdot r \ge \sqrt{p^2q^2+18pqr-27r^2-4q^3-4p^3r}\)
\(p=1 \) have; \((q-3r)-2f(q)\cdot r \ge \sqrt{q^2+18qr-27r^2-4q^3-4r}\)
\(\Leftrightarrow\)\((27+k^2)r^2+2(2-kq-9q)r+4q^3 \ge 0\)
\(\Delta_r ‘=(2-kq-9q)^2-4q^3(27+k^2) \)
\(=q^2(1-4q)k^2+2q(9q-2)k+(9q-2)^2-108q^3\)
Cho\(\Delta_r ‘=0 \) tìm dc \(k=\dfrac{2-9q\pm 4\sqrt{q(1-3q)^3}}{q(1-4q)}.\)
Ta chọn \(k=\dfrac{2-9q+ 4\sqrt{q(1-3q)^3}}{q(1-4q)}\). do đó \(f(q)=\dfrac{k-3}{2}=\dfrac{1-6q+6q^2+ 2\sqrt{q(1-3q)^3}}{q(1-4q)}\)
Suy ra
\( 1-6q+6q^2+ 2\sqrt{q(1-3q)^3}=\left[2\sqrt{q(1-3q)^3}-2(9q^2-2q)\right]+(24q^2-10q+1)\\ \)
\(=2\cdot \dfrac{q(1-3q)^3-(9q^2-2q)^2}{\sqrt{q(1-3q)^3}+2(9q^2-2q)}+(4q-1)(6q-1)\\ \)
\(=2\cdot \dfrac{q(1-4q)(27q^2-9q+1)}{\sqrt{q(1-3q)^3}+2(9q^2-2q)}+(4q-1)(6q-1)\)
Vậy \(f(q)=\dfrac{2(27q^2-9q+1)}{\sqrt{q(1-3q)^3}+2(9q^2-2q)}+\dfrac{1-6q}{q}\)
*Th1: Xét a;b < 0 thì \(a\le-2;b\le-2\)
khi đó VF âm và VT luôn dương nên BĐT luôn xảy ra.
*Th2: Xét a;b > 0 thì \(a\ge2;b\ge2\).
\(BDT\Leftrightarrow2a^2b^2+2a^2+2b^2+2\ge2\left(ab+1\right)\left(a+b\right)+10\)
\(\Leftrightarrow\left[\left(a+b\right)^2+a^2b^2-2ab\left(a+b\right)\right]+\left(a^2b^2-8ab+16\right)+\left(a^2+b^2-2ab\right)+8ab-2a-2b-24\ge0\)
\(\Leftrightarrow\left(a+b-ab\right)^2+\left(ab-4\right)^2+\left(a-b\right)^2+\left(a-2\right)\left(b-2\right)+7\left(ab-4\right)\ge0\)
( đúng)
Vậy BĐT được chứng minh.
tks