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Không mất tính tổng quát, giả sử \(a\ge b\ge c\)
\(\Rightarrow P\le\dfrac{a}{b+c+1}+\dfrac{b}{b+c+1}+\dfrac{c}{b+c+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)\)
\(\Rightarrow P\le\dfrac{a+b+c}{b+c+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)=\dfrac{a-1}{b+c+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)+1\)
\(\Rightarrow P\le\left(1-a\right)\left[\left(1-b\right)\left(1-c\right)-\dfrac{1}{b+c+1}\right]+1\le\left(1-a\right)\left[\left(1-b\right)\left(1-c\right)-\dfrac{1}{bc+b+c+1}\right]+1\)
\(\Rightarrow P\le\left(1-a\right)\left[\left(1-b\right)\left(1-c\right)-\dfrac{1}{\left(1+b\right)\left(1+c\right)}\right]+1\)
\(\Rightarrow P\le\left(1-a\right)\left(\dfrac{\left(1-b^2\right)\left(1-c^2\right)-1}{\left(1+b\right)\left(1+c\right)}\right)+1\)
Do \(a;b;c\le1\Rightarrow\left\{{}\begin{matrix}1-a\ge0\\\left(1-b^2\right)\left(1-c^2\right)\le1\\\end{matrix}\right.\) \(\Rightarrow\left(1-a\right)\left[\dfrac{\left(1-b^2\right)\left(1-c^2\right)-1}{\left(1+b\right)\left(1+c\right)}\right]\le0\)
\(\Rightarrow P\le1\)
\(P_{max}=1\) khi \(\left(a;b;c\right)=\left(0;0;0\right);\left(1;1;1\right);\left(0;1;1\right);\left(0;0;1\right)\) và các hoán vị
Đề bài hình như bị sai em, thay điểm rơi ko thỏa mãn
Biểu thức là \(a+b+\sqrt{2\left(a+c\right)}\) mới đúng
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+c}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c\left(b+c\right)}{8a^3\left(b+c\right)b^2c}}=\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{c+a}{4ca}+\dfrac{1}{2c}\ge3\sqrt[3]{\dfrac{c^2a\left(c+a\right)}{8b^3\left(c+a\right)c^2a}}=\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{a+b}{4ab}+\dfrac{1}{2a}\ge3\sqrt[3]{\dfrac{a^2b\left(a+b\right)}{8c^3\left(a+b\right)a^2b}}=\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{1}{4b}+\dfrac{1}{2b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{1}{4c}+\dfrac{1}{2c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{1}{4a}+\dfrac{1}{2a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{3}{4b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{3}{4c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{3}{4a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Rightarrow VT+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Đây là BĐT Iran 96 khá nổi tiếng. Bạn hoàn toàn có thể search trên google lời giải.
Đành giải tạm bằng nick này vì sợ một vài thành phần trẻ trâu anti phá phách :poor:
Phân tích và giải
Dễ thấy: Dấu "=" khi \(a=b=c=1\)
\(\Rightarrow L=Σ\dfrac{a}{\left(a+1\right)^2}=\dfrac{3}{4}\text{ và }F=-\dfrac{4}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}=-\dfrac{1}{2}\)
Khi đó \(VT=L-F=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\)
Ta sẽ chia làm 2 bước cm:
B1: \(Σ\dfrac{a}{\left(a+1\right)^2}\le\dfrac{3}{4}\). Ta xét BĐT :
\(\dfrac{a}{\left(a+1\right)^2}=\dfrac{a}{a^2+2a+1}\le\dfrac{3\left(a^{2k}+a^k\right)}{8\left(a^{2k}+a^k+1\right)}\) (cần tìm \(k\) thỏa mãn)
\(\Leftrightarrow8a\left(a^{2k}+a^k+1\right)-3\left(a^{2k}+a^k\right)\left(a^2+2a+1\right)\le0\)\(\Leftrightarrow f\left(a\right)=-3a^{2k}+2a^{k+1}-3a^{k+2}+2a^{2k+1}-3a^{2k+2}-3a^k+8a\)
\(\Rightarrow f'\left(a\right)=2k\cdot-3a^{2k-1}+\left(k+1\right)2a^k-\left(k+2\right)3a^{k+1}+\left(2k+1\right)2a^{2k}-\left(2k+2\right)3a^{2k+1}-k\cdot3a^{k-1}+8a\)
\(\Rightarrow f'\left(1\right)=0\Rightarrow-12k=0\Rightarrow k=0\)
Hay BĐT phụ cần tìm là \(\dfrac{a}{a^2+2a+1}\le\dfrac{3\left(a^{2\cdot0}+a^0\right)}{8\left(a^{2\cdot0}+a^0+1\right)}=\dfrac{1}{4}\) (bài này \(k\) đẹp ra luôn \(\farac{1}{4}\) cộng vào là ok =))
\(\Leftrightarrow-\dfrac{\left(a-1\right)^2}{4\left(a+1\right)^2}\le0\) *Đúng* \(\RightarrowΣ\dfrac{a}{\left(a+1\right)^2}\leΣ\dfrac{1}{4}=\dfrac{3}{4}\)
B2: CM \(-\dfrac{4}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\le-\dfrac{1}{2}\)
Tự cm nhé Goodluck :v
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 1
\(VT=\dfrac{a^2}{ab^2+abc+ac^2}+\dfrac{b^2}{c^2b+abc+a^2b}+\dfrac{c^2}{a^2c+abc+b^2c}\)
Áp dụng bđt Cauchy dạng phân thức
\(\Rightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b\right)+abc+ac\left(a+c\right)+abc+bc\left(b+c\right)+abc}\)
\(\Leftrightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}\)
\(\Leftrightarrow VT\ge\dfrac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)
Dấu ''='' xảy ra khi \(a=b=c\)
Bài 2
\(VT=\left(\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}\right)\left[\left(\dfrac{\sqrt{a}}{b+c}\right)^2+\left(\dfrac{\sqrt{b}}{c+a}\right)^2+\left(\dfrac{\sqrt{c}}{a+b}\right)^2\right]\)
Áp dụng bđt Bunhiacopxki ta có
\(VT\ge\left(\sqrt{a}.\dfrac{\sqrt{a}}{b+c}+\sqrt{b}.\dfrac{\sqrt{b}}{c+a}+\sqrt{c}.\dfrac{\sqrt{c}}{a+b}\right)^2\)
\(\Leftrightarrow VT\ge\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\)
Xét \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Áp dụng bđt Cauchy dạng phân thức ta có
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ac\right)}=\dfrac{3}{2}\)
\(\Rightarrow\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\ge\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)
\(\Rightarrow VT\ge\dfrac{9}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c\)
có mùi schur
:v