Sentence examples for using the last inequality from inspiring English sources

In successively using the last inequality, we deduce that {u}_s^{t+1}>{u}_s^t.

Using the last inequality (2.12) in (2.11), we get the desired inequality.

Using the last inequality in (3.20), we see that { w n } is a positive solution of △ ( w n + p β w τ ( n ) ) + Q n ∗ w σ ( n ) β α ≤ 0, which is a contradiction.

Using the last inequality and the fact that φ is non-decreasing, we have max ( d ( g x, g u n ), d ( g y, g v n ) ) ≤ max ( d ( g x, g u n − 1 ), d ( g y, g v n − 1 ) ).

Using the last two inequalities, we get begin{aligned} biglVert x t) bigrVert &leq bigl[ mu_{h}^{-1}V_{h} bigl(x theta_{i}+) bigr)exp bigl alpha_{h} t- theta_{i}) bigr) bigr]^{1/p} &leqsqrt[p]{mu_{h}^{-1}lambda_{h}}exp bigl( alpha_{h} t-theta_{i})/p bigr) biglVert x( theta_{i} t-theta_{i}

where Young's inequality,, has been used at the last inequality.

where (3.30) has been used in the last inequality and B i ′ ≡ B i + ( r + H i ) T 1 p i ∥ μ r, i ∥ q i.

end{aligned} Taking the supremum on both sides leads to |Sx|geqlambdaint_{z}^{1-eta} G(t_{z},s a(s) (s-eta)^{m-q-2},dscdot underline{f}_{infty}-epsilon)frac{gamma|x|}{ m-q-2)!}geq|x|, where (t_{ m-q-2 defined in (4.1), and (4.3) has been used in the last inequality.

where in the second last inequality a similar argument as in (3.34) is used, and in the last inequality we have used (3.37).

Using (3.40) and the last inequality, we obtain that (3.61).

Letting k → ∞ and using (2.7) in the last inequality, we have lim k → ∞ r k = lim k → ∞ [ d ( g x n ( k ), g x m ( k ) ) + d ( g y n ( k ), g y m ( k ) ) 2 ] = ε.