Cho a,b>0 thoả mãn: a+b=1. CMR: \(\dfrac{2}{ab}+\dfrac{3}{a^2+b^2}\ge14\)
Cho các số dương \(a,b\) thỏa mãn \(a+b=1\). Chứng minh: \(\dfrac{2}{ab}+\dfrac{3}{a^2+b^2}\ge14\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
$\frac{2}{ab}+\frac{3}{a^2+b^2}=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{1}{a^2+b^2}+\frac{1}{a^2+b^2}$
$\geq \frac{(1+1+1+1+1+1+1)^2}{2ab+2ab+2ab+2ab+a^2+b^2+a^2+b^2+a^2+b^2}=\frac{49}{8ab+3(a^2+b^2)}$
$=\frac{49}{3(a+b)^2+2ab}\geq \frac{49}{3(a+b)^2+\frac{(a+b)^2}{2}}=\frac{49}{3+\frac{1}{2}}=14$
Ta có đpcm
Dấu "=" xảy ra khi $a=b=\frac{1}{2}$
Cho các số thực dương a,b và c thoả mãn: \(\dfrac{1}{a+2}\)+\(\dfrac{1}{b+2}\)+\(\dfrac{1}{c+2}\)\(\ge\dfrac{3}{2}\)
CMR: \(a+b+c\ge ab+bc+ca\)
\(\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}\ge2\)
\(\Leftrightarrow\dfrac{2}{a+2}-1+\dfrac{2}{b+2}-1+\dfrac{2}{c+2}-1\ge2-3\)
\(\Rightarrow1\ge\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}=\dfrac{a^2}{a^2+2a}+\dfrac{b^2}{b^2+2b}+\dfrac{c^2}{c^2+2c}\)
\(\Rightarrow1\ge\dfrac{\left(a+b+c\right)^2}{a^2+2a+b^2+2b+c^2+2c}\)
\(\Rightarrow a^2+b^2+c^2+2\left(a+b+c\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Rightarrow\) đpcm
Phía trên thoả mãn \(\ge1\) chứ không phải 3/2 đâu ạ
Cho các số thực dương thoả mãn:
\(\dfrac{1}{a+2}\) + \(\dfrac{1}{b+2}\)+ \(\dfrac{1}{c+2}\)\(\ge1\). CMR: \(a+b+c\ge ab+bc+ca\)
1) Cho a, b, c ∈ [0;1] và a + b + c = 2. CMR ab + bc + ca ≥ 2abc + \(\dfrac{20}{27}\)
2) Cho a, b, c ∈ [1;3] và a + b + c = 6. CMR a3 + b3 + c3 ≤ 36
3) Cho các số dương a, b, c, d thoả mãn a + b + c + d = 4. CMR \(\dfrac{a}{1+b^2}+\dfrac{b}{1+c^2}+\dfrac{c}{1+d^2}+\dfrac{d}{1+a^2}\) ≥ 2
1.
Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có 2 số cùng phía so với \(\dfrac{2}{3}\), không mất tính tổng quát, giả sử đó là b và c
\(\Rightarrow\left(b-\dfrac{2}{3}\right)\left(c-\dfrac{2}{3}\right)\ge0\)
Mặt khác \(0\le a\le1\Rightarrow1-a\ge0\)
\(\Rightarrow\left(b-\dfrac{2}{3}\right)\left(c-\dfrac{2}{3}\right)\left(1-a\right)\ge0\)
\(\Leftrightarrow-abc\ge\dfrac{4a}{9}+\dfrac{2b}{3}+\dfrac{2c}{3}-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc-\dfrac{4}{9}\)
\(\Leftrightarrow-abc\ge-\dfrac{2a}{9}+\dfrac{2}{3}\left(a+b+c\right)-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc-\dfrac{4}{9}=-\dfrac{2a}{9}-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc+\dfrac{8}{9}\)
\(\Leftrightarrow-2abc\ge-\dfrac{4a}{9}-\dfrac{4ab}{3}-\dfrac{4ac}{3}-2bc+\dfrac{16}{9}\)
\(\Leftrightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}-\dfrac{ab}{3}-\dfrac{ac}{3}-bc+\dfrac{16}{9}\)
\(\Leftrightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}-\dfrac{a}{3}\left(b+c\right)-bc+\dfrac{16}{9}\ge-\dfrac{4a}{9}-\dfrac{a}{3}\left(2-a\right)-\dfrac{\left(b+c\right)^2}{4}+\dfrac{16}{9}\)
\(\Rightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}+\dfrac{a^2}{3}-\dfrac{2a}{3}-\dfrac{\left(2-a\right)^2}{4}+\dfrac{16}{9}\)
\(\Rightarrow ab+bc+ca-2abc\ge\dfrac{a^2}{12}-\dfrac{a}{9}+\dfrac{7}{9}=\dfrac{1}{12}\left(a-\dfrac{2}{3}\right)^2+\dfrac{20}{27}\ge\dfrac{20}{27}\)
\(\Rightarrow ab+bc+ca\ge2abc+\dfrac{20}{27}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\)
2.
Đặt \(\left(a;b;c\right)=\left(x+1;y+1;z+1\right)\Rightarrow\left\{{}\begin{matrix}x;y;z\in\left[0;2\right]\\x+y+z=3\end{matrix}\right.\)
Ta có: \(P=\left(x+1\right)^3+\left(y+1\right)^3+\left(z+1\right)^3\)
\(P=x^3+y^3+z^3+3\left(x^2+y^2+z^2\right)+12\)
Không mất tính tổng quát, giả sử \(x\ge y\ge z\Rightarrow x\ge1\)
\(\Rightarrow\left\{{}\begin{matrix}y^3+z^3=\left(y+z\right)^3-3yz\left(y+z\right)\le\left(y+z\right)^3\\y^2+z^2=\left(y+z\right)^2-2yz\le\left(y+z\right)^2\end{matrix}\right.\)
\(\Rightarrow P\le x^3+\left(3-x\right)^3+3x^2+3\left(3-x\right)^2+12\)
\(\Rightarrow P\le15x^2-45x+66=15\left(x-1\right)\left(x-2\right)+36\le36\)
(Do \(1\le x\le2\Rightarrow\left(x-1\right)\left(x-2\right)\le0\))
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(2;1;0\right)\) và các hoán vị hay \(\left(a;b;c\right)=\left(1;2;3\right)\) và các hoán vị
Cho a,b,c >0 thoả mãn ab+bc+ca=3. Tìm giá trị nhỏ nhất của
P=\(\dfrac{1+3a}{1+b^2}+\dfrac{1+3b}{1+c^2}+\dfrac{1+3c}{1+a^2}\)
Ta có BĐT: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=3.3=9\)
\(\Rightarrow a+b+c\ge3\)
Phân tích và áp dụng BĐT AM-GM:
\(\dfrac{1+3a}{1+b^2}=\dfrac{1}{1+b^2}+\dfrac{3a}{1+b^2}=\left(1-\dfrac{b^2}{1+b^2}\right)+\left(3a-\dfrac{3ab^2}{1+b^2}\right)\ge\left(1-\dfrac{b^2}{2b}\right)+\left(3a-\dfrac{3ab^2}{2b}\right)=\left(1-\dfrac{b}{2}\right)+\left(3a-\dfrac{3}{2}ab\right)\)
Tương tự:
\(\dfrac{1+3b}{1+c^2}\ge\left(1-\dfrac{c}{2}\right)+\left(3b-\dfrac{3}{2}bc\right)\)
\(\dfrac{1+3c}{1+a^2}\ge\left(1-\dfrac{a}{2}\right)+\left(3c-\dfrac{3}{2}ca\right)\)
Cộng các vế của các BĐT ta được:
\(P\ge3-\dfrac{1}{2}\left(a+b+c\right)+3\left(a+b+c\right)-\dfrac{3}{2}\left(ab+bc+ca\right)=3+\dfrac{5}{2}\left(a+b+c\right)-\dfrac{3}{2}.3\ge3+\dfrac{5}{2}.3-\dfrac{9}{2}=6\)
\(P=6\Leftrightarrow a=b=c=1\)
Vậy \(P_{min}=6\)
Cho 3 số a,b,c thỏa mãn ab + bc + ca = 1. CMR:
\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)
Đặt A = \(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)
= \(\dfrac{a-b}{c^2+ab+bc+ca}+\dfrac{b-c}{a^2+ab+bc+ca}+\dfrac{c-a}{b^2+ab+bc+ca}\)
= \(\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(c+a\right)}+\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}\)
= \(\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c+a\right)\left(c-a\right)}{\left(c+a\right)\left(b+c\right)\left(a+b\right)}\)
= \(\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}\)
\(=\dfrac{a-b}{ab+bc+ca+c^2}+\dfrac{b-c}{ab+bc+ca+a^2}+\dfrac{c-a}{ab+bc+ca+b^2}\)
\(=\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(a+c\right)}+\dfrac{c-a}{\left(b+a\right)\left(b+c\right)}\)
\(=\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
cho 3 số a, b, c thoả mãn 0 < a, b, c < 1.CMR
\(\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{3}{3+abc}\)
Cho các số thực dương a, b, c thoả mãn:
\(a\sqrt{1-b^2}+b\sqrt{1-c^2}+c\sqrt{1-a^2}=\dfrac{3}{2}\)
Cmr: \(a^2+b^2+c^2=\dfrac{3}{2}\)
Cho 2 số nguyên dường \(a\) và \(b\) thoả mãn \(\dfrac{3}{25}< \dfrac{a}{b}< \dfrac{4}{33}\)
CMR : \(b>=58\)