Em kiểm tra lại mẫu số của biểu thức c, chắc chắn đề sai

Chia 2 vế cho \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\) BĐT trở thành:

\(\dfrac{1}{a^4\left(b+1\right)\left(c+1\right)}+\dfrac{1}{b^4\left(a+1\right)\left(c+1\right)}+\dfrac{1}{c^4\left(a+1\right)\left(b+1\right)}\ge\dfrac{3}{4}\)

Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\) \(\Rightarrow xyz=1\)

\(\dfrac{1}{a^4\left(b+1\right)\left(c+1\right)}=\dfrac{x^4}{\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)}=\dfrac{x^4yz}{\left(y+1\right)\left(z+1\right)}=\dfrac{x^3}{\left(y+1\right)\left(z+1\right)}\)

Do đó BĐT trở thành:

\(\dfrac{x^3}{\left(y+1\right)\left(z+1\right)}+\dfrac{y^3}{\left(x+1\right)\left(z+1\right)}+\dfrac{z^3}{\left(x+1\right)\left(y+1\right)}\ge\dfrac{3}{4}\)

Một bài toán quen thuộc

Trước hết ta c/m bổ đề sau:

Với mọi số thực dương x;y ta luôn có:

\(x^4+y^4\ge xy\left(x^2+y^2\right)\)

Thật vậy, BĐT đã cho tương đương:

\(x^4-x^3y+y^4-xy^3\ge0\Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)\left(x^3-y^3\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\) (luôn đúng)

Áp dụng bổ đề trên ta có:

\(T\le\dfrac{a}{bc\left(b^2+c^2\right)+a}+\dfrac{b}{ac\left(a^2+c^2\right)+b}+\dfrac{c}{ab\left(a^2+b^2\right)+c}\)

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

\(\Rightarrow T\le\dfrac{a^2}{a^2+b^2+c^2}+\dfrac{b^2}{a^2+b^2+c^2}+\dfrac{c^2}{a^2+b^2+c^2}=1\)

\(T_{max}=1\) khi \(a=b=c=1\)

Với mọi số thực dương a;b;c ta có BĐT:

\(a^4+b^4\ge ab\left(a^2+b^2\right)\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

Tương tự, ta có:

\(VT\le\dfrac{ab}{ab\left(a^2+b^2\right)+ab}+\dfrac{bc}{bc\left(b^2+c^2\right)+bc}+\dfrac{ca}{ca\left(c^2+a^2\right)+ca}\)

\(VT\le\dfrac{1}{a^2+b^2+1}+\dfrac{1}{b^2+c^2+1}+\dfrac{1}{c^2+a^2+1}\)

Đặt \(\left(a^2;b^2;c^2\right)=\left(x^3;y^3;z^3\right)\Rightarrow xyz=1\)

\(VT\le\dfrac{1}{x^3+y^3+1}+\dfrac{1}{y^3+z^3+1}+\dfrac{1}{z^3+x^3+1}\)

Ta lại có: \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)

\(\Rightarrow VT\le\dfrac{xyz}{xy\left(x+y\right)+xyz}+\dfrac{xyz}{yz\left(y+z\right)+xyz}+\dfrac{xyz}{zx\left(z+x\right)+xyz}=1\)

Nhân vế với vế của giả thiết:

\(\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{c}\right)\left(c+\dfrac{1}{a}\right)=\dfrac{28}{3}\)

\(\Leftrightarrow\left(ab+\dfrac{1}{bc}+\dfrac{a}{c}+1\right)\left(c+\dfrac{1}{a}\right)=\dfrac{28}{3}\)

\(\Leftrightarrow abc+\dfrac{1}{abc}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+a+b+c=\dfrac{28}{3}\) (1)

Cộng vế với vế giả thiết:

\(\Rightarrow a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=4+1+\dfrac{7}{3}=\dfrac{22}{3}\) (2)

(1);(2) \(\Rightarrow abc+\dfrac{1}{abc}+\dfrac{22}{3}=\dfrac{28}{3}\)

\(\Rightarrow abc+\dfrac{1}{abc}=2\)

\(\Rightarrow\left(abc\right)^2-2\left(abc\right)+1=0\)

\(\Rightarrow\left(abc-1\right)^2=0\)

\(\Rightarrow abc=1\)

Theo giả thiết kết hợp sử dụng BĐT AM - GM có:

\(\left(a+b-c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}\right)=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+1-\left[c\left(a+b\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]\)

\(\le\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+1-2\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}=\left[\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1\right]^2\)

Suy ra \(\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1\ge2\Leftrightarrow\sqrt{\dfrac{a}{b}+\dfrac{b}{a}+2}\ge3\)

\(\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge7\)

Khi đó, sử dụng BĐT Cauchy - Schwarz ta có:

\(\left(a^4+b^4+c^4\right)\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}\right)\ge\left[\sqrt{\left(a^4+b^4\right)\left(\dfrac{1}{a^4}+\dfrac{1}{b^4}\right)}+1\right]^2\)

\(=\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+1\right)^2=\left[\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2-1\right]^2\ge\left(7^2-1\right)^2=2304\)

Đẳng thức xảy ra khi và chỉ khi \(ab=c^2\) và \(\dfrac{a}{b}+\dfrac{b}{a}=7\)

(a+b-c)(1/a+1/b-c)=(a+b)(1/a+1/b)+1-[c(a+b)+c(1/a+1/b)]<=(a+b)(1/a+1/b)+1-2căn (a+b)(1/a+1/b)

=[(căn (a+b)(1/a+1/b))-1]^2

=>\(\sqrt{\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}-1>=2\)

=>\(\sqrt{\dfrac{a}{b}+\dfrac{b}{a}+2}>=3\)

=>a/b+b/a>=7

(a^4+b^4+c^4)(1/a^4+1/b^4+1/c^4)>=[căn ((a^4+b^4)(1/a^4+1/b^4))+1]^2

=(a^2/b^2+b^2/a^2+1)^2=[(a/b+b/a)^2-1]^2>=(7^2-1)^2=2304

=>ĐPCM

ôn zăn ik bà, tón gánh hong nổi đâu =))

Áp dụng BĐT BSC:

\(P=\dfrac{1}{a}+\dfrac{4}{b}+\dfrac{9}{c}\ge\dfrac{\left(1+2+3\right)^2}{a+b+c}=\dfrac{36}{1}=36\)

\(minP=36\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{6}\\b=\dfrac{1}{3}\\c=\dfrac{1}{2}\end{matrix}\right.\)